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2 <p>Last updated on<strong>November 27, 2025</strong></p>

3 <p>Conditional probability is when an event occurs only if certain conditions are met. This concept can be used to determine the likelihood of real-life events occurring under specific conditions. For example, in weather forecasting, we can calculate the probability of rain. In this topic, students will learn how conditional probability can be applied easily.</p>

4 <h2>What is Conditional Probability?</h2>

5 <p>Conditional<a>probability</a>is an essential idea in<a>probability and statistics</a>. It tells us the chance that event A occurs after we already know that event B has occurred. We can write this as P(A | B), P(A / B), or PB(A). In simple words, P(A | B) means “the probability of A occurring given that B has happened.” Because B gives us new information, the probability of A may change.</p>

6 <p>This idea is often explained with a conditional probability example: knowing one condition affects the final answer.</p>

7 <p><strong>Example:</strong>A card is drawn from a standard deck of 52 cards. If it is known that the card is red, find the probability that the card is a king.</p>

8 <p><strong>Solution:</strong></p>

9 <p>Let: A = event that the card drawn is a king B = event that the card drawn is red</p>

10 <p>We want that: P(A∣B)</p>

11 <p><strong>Step 1:</strong>First, we need to determine the total<a>number</a>of red cards (event B). There are 26 red cards (13 hearts + 13 diamonds).</p>

12 <p><strong>Step 2:</strong>Next, we find the number of red kings (A ∩ B). There are two red kings (the King of Hearts and the King of Diamonds).</p>

13 <p><strong>Step 3:</strong>Then apply the conditional probability<a>formula</a>: P(A∣B) = \(\frac{\text{Number of red kings}}{\text{Total red cards}} = \frac{2}{26} = \frac{1}{13}\)</p>

14 <h2>Conditional Probability Formula</h2>

15 <p>In the card example, we found that: P(A∣B) = \(\frac{1}{13}\)</p>

16 <p>Here:</p>

17 <p>The 1 represents the one specific event that satisfies both A and B. The card drawn is a red king. The 13 represents the total number of possible outcomes in event B. There are 13 red cards in each suit, but we treat the total as 26, so the simplified probability becomes 113. This reasoning helps us to understand how the conditional probability<a>equation</a>is formed.</p>

18 <p><strong>General Formula</strong></p>

19 <p>Probability of A given B: P(A∣B) = P(A∩B)P(B), (where P(B) = 0)</p>

20 <p>Probability of B given A: P(B∣A) = P(A∩B)P(A), (where P(A) = 0).</p>

21 <p>These formulas are part of the Kolmogorov definition of conditional probability.</p>

22 <p>Here is the meaning of each<a>term</a>:</p>

23 <ul><li>P(A | B): Probability of drawing the king, given that the card drawn is red. </li>

24 <li>P(B | A): Probability that the card is red, given that it is the king. </li>

25 <li>P(A ∩ B): Probability that the card drawn is both red and the king, which means a red king. </li>

26 <li>P(A): Probability of drawing a king from the deck. </li>

27 <li>P(B): Probability of drawing a red card from the deck.</li>

28 </ul><h2>Difference between Conditional Probability, Joint Probability and Marginal Probability</h2>

29 <p>The three types of probability are conditional, joint, and marginal. Here, we will discuss a few key points that differentiate them: </p>

30 <strong>Features </strong><strong>Conditional Probability</strong><strong>Joint Probability</strong><strong>Marginal Probability</strong>Meaning The probability of event A happening, given that event B has been already occurred. Formula: P(A∣B) = P(A∩B)P(B) The probability that two events occur at the same time, A and B. Formula: P(A∩B) The probability that an event will occur, regardless of other events. Formula \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) P(A∩B) \(P(A) = \sum P(A \cap B)\) Variables Involved Two or more related events Two or more events A single event<h3>Explore Our Programs</h3>

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32 <h2>How to Find Conditional Probability?</h2>

31 <h2>How to Find Conditional Probability?</h2>

33 <p>To find the conditional probability, we use the following steps:</p>

32 <p>To find the conditional probability, we use the following steps:</p>

34 <p>1. Identify the events</p>

33 <p>1. Identify the events</p>

35 <ul><li>The event whose probability needs to be determined (A). </li>

34 <ul><li>The event whose probability needs to be determined (A). </li>

36 <li>An event that has already occurred (B).</li>

35 <li>An event that has already occurred (B).</li>

37 </ul><p>2. Calculate the joint probability P(A ∩ B) </p>

36 </ul><p>2. Calculate the joint probability P(A ∩ B) </p>

38 <ul><li>The probability of both events occurring at the same time.</li>

37 <ul><li>The probability of both events occurring at the same time.</li>

39 </ul><p>3. Determine the probability of the stated condition P(B).</p>

38 </ul><p>3. Determine the probability of the stated condition P(B).</p>

40 <p>4. Now, we use the formula: P(A|B) = P(A ∩ B) / P(B)</p>

39 <p>4. Now, we use the formula: P(A|B) = P(A ∩ B) / P(B)</p>

41 <h3><strong>Conditional Probability for Independent Events</strong></h3>

40 <h3><strong>Conditional Probability for Independent Events</strong></h3>

42 <p>The probability of<a>independent events</a>calculates the probability of events whose occurrences do not depend on each other. Here, the conditional probability of two independent events, A and B, can be written as:</p>

41 <p>The probability of<a>independent events</a>calculates the probability of events whose occurrences do not depend on each other. Here, the conditional probability of two independent events, A and B, can be written as:</p>

43 <p>P(A|B) = P(A)</p>

42 <p>P(A|B) = P(A)</p>

44 <p>P(B|A) = P(B)</p>

43 <p>P(B|A) = P(B)</p>

45 <p>These events satisfy the <a>multiplication</a>rule: P(A ∩ B) = P(A) × P(B)</p>

44 <p>These events satisfy the <a>multiplication</a>rule: P(A ∩ B) = P(A) × P(B)</p>

46 <h3><strong>Conditional Probability for Mutually Exclusive Events</strong></h3>

45 <h3><strong>Conditional Probability for Mutually Exclusive Events</strong></h3>

47 <p>Mutually exclusive events are those that cannot occur simultaneously. For<a>mutually exclusive events</a>, the conditional probability will always be equal to zero.</p>

46 <p>Mutually exclusive events are those that cannot occur simultaneously. For<a>mutually exclusive events</a>, the conditional probability will always be equal to zero.</p>

48 <p>P(A|B) = 0</p>

47 <p>P(A|B) = 0</p>

49 <p>P(B|A) = 0</p>

48 <p>P(B|A) = 0</p>

50 <h2>Conditional Probability and Bayes’ Theorem</h2>

49 <h2>Conditional Probability and Bayes’ Theorem</h2>

51 <p>Bayes’ theorem helps us to find the probability of an event when we have new information about another related event. It allows us to update or revise our earlier probabilities based on this new evidence. In simple terms, Bayes’ theorem tells us how likely a cause is, given that we have observed a particular result. </p>

50 <p>Bayes’ theorem helps us to find the probability of an event when we have new information about another related event. It allows us to update or revise our earlier probabilities based on this new evidence. In simple terms, Bayes’ theorem tells us how likely a cause is, given that we have observed a particular result. </p>

52 <p>A tree diagram is used to understand the conditional probabilities. It visually shows each possible outcome step by step, helping us to see how the probabilities split and change at every stage.</p>

51 <p>A tree diagram is used to understand the conditional probabilities. It visually shows each possible outcome step by step, helping us to see how the probabilities split and change at every stage.</p>

53 <h2>Properties of Conditional Probabilitiy</h2>

52 <h2>Properties of Conditional Probabilitiy</h2>

54 <p>Conditional probability has a few rules that help us understand how probabilities change when we already know that one event has occurred. These rules make it easier to calculate and compare the probabilities. Here are some properties that are explained in simple terms:</p>

53 <p>Conditional probability has a few rules that help us understand how probabilities change when we already know that one event has occurred. These rules make it easier to calculate and compare the probabilities. Here are some properties that are explained in simple terms:</p>

55 <p><strong> Property 1:</strong>Whole Sample Space Has Probability 1 If E and F are events in the<a>sample space</a>S, then: P(S∣F) = P(F∣F) = 1</p>

54 <p><strong> Property 1:</strong>Whole Sample Space Has Probability 1 If E and F are events in the<a>sample space</a>S, then: P(S∣F) = P(F∣F) = 1</p>

56 <p>Meaning: If we already know that F has happened, then we are completely sure, 100% certain, that the outcome lies in F. So the probability is 1.</p>

55 <p>Meaning: If we already know that F has happened, then we are completely sure, 100% certain, that the outcome lies in F. So the probability is 1.</p>

57 <p><strong> Property 2:</strong>Conditional Probability of A or B If A, B, and F are events, and P(F) ≠ 0, then: P(A∪B∣F) = P(A∣F) + P(B∣F) - P(A∩B∣F)</p>

56 <p><strong> Property 2:</strong>Conditional Probability of A or B If A, B, and F are events, and P(F) ≠ 0, then: P(A∪B∣F) = P(A∣F) + P(B∣F) - P(A∩B∣F)</p>

58 <p>Meaning: When we want the probability of A or B happening under the condition F, we: Add the conditional probabilities of A and B. Subtract the part where both A and B occur together (to avoid double-counting). This rule is the same as the usual<a>addition</a>rule, just applied under the condition F.</p>

57 <p>Meaning: When we want the probability of A or B happening under the condition F, we: Add the conditional probabilities of A and B. Subtract the part where both A and B occur together (to avoid double-counting). This rule is the same as the usual<a>addition</a>rule, just applied under the condition F.</p>

59 <p><strong> Property 3:</strong>Conditional Probability of Complements</p>

58 <p><strong> Property 3:</strong>Conditional Probability of Complements</p>

60 <p> The probability of A not happening, given B, is: P(A′∣B) = 1 - P(A∣B)</p>

59 <p> The probability of A not happening, given B, is: P(A′∣B) = 1 - P(A∣B)</p>

61 <p>Meaning: Once B has happened, only two outcomes are possible: A happens, or A does not occur.</p>

60 <p>Meaning: Once B has happened, only two outcomes are possible: A happens, or A does not occur.</p>

62 <p>Since these two cover all possibilities under B, their conditional probabilities must add up to 1. So we can easily find one by subtracting the other from 1.</p>

61 <p>Since these two cover all possibilities under B, their conditional probabilities must add up to 1. So we can easily find one by subtracting the other from 1.</p>

63 <h2>Tips and Tricks For Conditional Probability</h2>

62 <h2>Tips and Tricks For Conditional Probability</h2>

64 <p>Understanding conditional probability becomes much easier when you know a few smart strategies. These tips and tricks help you break problems into simpler parts, avoid common mistakes, and solve<a>questions</a>faster and more accurately.</p>

63 <p>Understanding conditional probability becomes much easier when you know a few smart strategies. These tips and tricks help you break problems into simpler parts, avoid common mistakes, and solve<a>questions</a>faster and more accurately.</p>

65 <ul><li>Encourage students to use visual tools such as trees and Venn diagrams. </li>

64 <ul><li>Encourage students to use visual tools such as trees and Venn diagrams. </li>

66 <li>Give them small, real-life examples like weather, tests, or games to make concepts relatable. </li>

65 <li>Give them small, real-life examples like weather, tests, or games to make concepts relatable. </li>

67 <li>Provide the conditional probability<a>worksheets</a>so they can practice step-by-step. </li>

66 <li>Provide the conditional probability<a>worksheets</a>so they can practice step-by-step. </li>

68 <li>Use a conditional probability<a>calculator</a>to verify the answers, and it will build confidence. </li>

67 <li>Use a conditional probability<a>calculator</a>to verify the answers, and it will build confidence. </li>

69 <li>Whenever you see the words “given that,” it indicates that an event has already happened; this event becomes your new sample space. </li>

68 <li>Whenever you see the words “given that,” it indicates that an event has already happened; this event becomes your new sample space. </li>

70 <li>If you are given B, ignore everything outside B. Your total now becomes P(B), not the whole sample space. </li>

69 <li>If you are given B, ignore everything outside B. Your total now becomes P(B), not the whole sample space. </li>

71 <li>Do Not Mix Up P(A | B) and P(B | A). These two are different, and confusing them is a common mistake. Always check which one the question is asking for.</li>

70 <li>Do Not Mix Up P(A | B) and P(B | A). These two are different, and confusing them is a common mistake. Always check which one the question is asking for.</li>

72 </ul><h2>Common Mistakes and How to Avoid Them in Conditional Probability</h2>

71 </ul><h2>Common Mistakes and How to Avoid Them in Conditional Probability</h2>

73 <p>Conditional probability is a significant concept in probability theory. Students can often make mistakes when determining conditional probability. These errors can be avoided with a proper understanding of the concept. Here are a few common mistakes along with some tricks to help you avoid them.</p>

72 <p>Conditional probability is a significant concept in probability theory. Students can often make mistakes when determining conditional probability. These errors can be avoided with a proper understanding of the concept. Here are a few common mistakes along with some tricks to help you avoid them.</p>

74 <h2>Real-Life Applications of Conditional Probability</h2>

73 <h2>Real-Life Applications of Conditional Probability</h2>

75 <p>Conditional probability plays a vital role in the prediction of various real-life situations based on conditions. Whether predicting the weather or solving complex real-life problems, conditional probability has numerous uses. Let’s look at a few examples:</p>

74 <p>Conditional probability plays a vital role in the prediction of various real-life situations based on conditions. Whether predicting the weather or solving complex real-life problems, conditional probability has numerous uses. Let’s look at a few examples:</p>

76 <ul><li>Conditional probability is widely used in weather forecasting by predicting the probability of rain based on the condition that the sky is cloudy.</li>

75 <ul><li>Conditional probability is widely used in weather forecasting by predicting the probability of rain based on the condition that the sky is cloudy.</li>

77 </ul><ul><li>In healthcare, the probability of confirming a disease if the test is positive.</li>

76 </ul><ul><li>In healthcare, the probability of confirming a disease if the test is positive.</li>

78 </ul><ul><li>In traffic control, the probability of occurrence of accidents if the roads are overcrowded.</li>

77 </ul><ul><li>In traffic control, the probability of occurrence of accidents if the roads are overcrowded.</li>

79 </ul><ul><li>Students use conditional probability to predict the chance of failing an exam under certain conditions, such as a strict evaluation.</li>

78 </ul><ul><li>Students use conditional probability to predict the chance of failing an exam under certain conditions, such as a strict evaluation.</li>

80 </ul><ul><li>We use it in predicting the stock market returns. If the stocks are down, the probability of gaining returns will be low.</li>

79 </ul><ul><li>We use it in predicting the stock market returns. If the stocks are down, the probability of gaining returns will be low.</li>

81 </ul><h3>Problem 1</h3>

80 </ul><h3>Problem 1</h3>

82 <p>Aden has 60% chance of qualifying for an exam if they prepare and 30% chance if they don’t. If the probability of studying is 80%, what is the probability that Aden qualifies for the exam?</p>

81 <p>Aden has 60% chance of qualifying for an exam if they prepare and 30% chance if they don’t. If the probability of studying is 80%, what is the probability that Aden qualifies for the exam?</p>

83 <p>Okay, lets begin</p>

82 <p>Okay, lets begin</p>

84 <p>The probability of Aden qualifying for the exam is 54% or 0.54.</p>

83 <p>The probability of Aden qualifying for the exam is 54% or 0.54.</p>

85 <h3>Explanation</h3>

84 <h3>Explanation</h3>

86 <p>We have: </p>

85 <p>We have: </p>

87 <p>A = qualifying for the exam</p>

86 <p>A = qualifying for the exam</p>

88 <p>B = Preparing for the exam</p>

87 <p>B = Preparing for the exam</p>

89 <p><strong>The given probabilities are:</strong></p>

88 <p><strong>The given probabilities are:</strong></p>

90 <p>Probability of qualifying for the exam if they prepare</p>

89 <p>Probability of qualifying for the exam if they prepare</p>

91 <p>P(A|B) = 0.6 </p>

90 <p>P(A|B) = 0.6 </p>

92 <p>Probability of qualifying even without preparation:</p>

91 <p>Probability of qualifying even without preparation:</p>

93 <p>P(A|Bc) = 0.3</p>

92 <p>P(A|Bc) = 0.3</p>

94 <p>Probability of preparing for the exam:</p>

93 <p>Probability of preparing for the exam:</p>

95 <p>P(B) = 0.8</p>

94 <p>P(B) = 0.8</p>

96 <p>Probability of not preparing for the exam:</p>

95 <p>Probability of not preparing for the exam:</p>

97 <p>\(P(Bc) = 1 - P(B) = 1 - 0.8 = 0.2\)</p>

96 <p>\(P(Bc) = 1 - P(B) = 1 - 0.8 = 0.2\)</p>

98 <p>Here, we use the law of total probability:</p>

97 <p>Here, we use the law of total probability:</p>

99 <p>\(P(A) = P(A \mid B)\,P(B) + P(A \mid B^{c})\,P(B^{c})\)</p>

98 <p>\(P(A) = P(A \mid B)\,P(B) + P(A \mid B^{c})\,P(B^{c})\)</p>

100 <p>Now, substitute the values into the equation:</p>

99 <p>Now, substitute the values into the equation:</p>

101 <p>\(P(A) = (0.6 × 0.8) + (0.3 × 0.2)\)</p>

100 <p>\(P(A) = (0.6 × 0.8) + (0.3 × 0.2)\)</p>

102 <p>\(= 0.48 + 0.06\)</p>

101 <p>\(= 0.48 + 0.06\)</p>

103 <p>\(= 0.54\)</p>

102 <p>\(= 0.54\)</p>

104 <p>Therefore, we get 0.54 as the probability of Aden qualifying for the exam, which is equal to 54%.</p>

103 <p>Therefore, we get 0.54 as the probability of Aden qualifying for the exam, which is equal to 54%.</p>

105 <p>Well explained 👍</p>

104 <p>Well explained 👍</p>

106 <h3>Problem 2</h3>

105 <h3>Problem 2</h3>

107 <p>In a class of 80 pupils, 50 learn Mathematics, 30 learn English, and 15 learn both Mathematics and English. If a pupil is selected randomly and is found to be learning English, what is the probability that they also learn Mathematics?</p>

106 <p>In a class of 80 pupils, 50 learn Mathematics, 30 learn English, and 15 learn both Mathematics and English. If a pupil is selected randomly and is found to be learning English, what is the probability that they also learn Mathematics?</p>

108 <p>Okay, lets begin</p>

107 <p>Okay, lets begin</p>

109 <p>The probability of a pupil learning both English and Math at the same time is 0.5 or 50%.</p>

108 <p>The probability of a pupil learning both English and Math at the same time is 0.5 or 50%.</p>

110 <h3>Explanation</h3>

109 <h3>Explanation</h3>

111 <p>Here,</p>

110 <p>Here,</p>

112 <p>A = Pupil learning mathematics</p>

111 <p>A = Pupil learning mathematics</p>

113 <p>B = Pupil learning English</p>

112 <p>B = Pupil learning English</p>

114 <p>Now, we determine P(A|B):</p>

113 <p>Now, we determine P(A|B):</p>

115 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

114 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

116 <p>We will now change the given values into probabilities:</p>

115 <p>We will now change the given values into probabilities:</p>

117 <p>\(P(A \cap B) = \frac{15}{80}\)</p>

116 <p>\(P(A \cap B) = \frac{15}{80}\)</p>

118 <p>\(P(B) = \frac{30}{80}\)</p>

117 <p>\(P(B) = \frac{30}{80}\)</p>

119 <p>Using the formula: \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

118 <p>Using the formula: \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

120 <p>= \(\frac{\frac{15}{80}}{\frac{30}{80}} = \frac{15}{30} = 0.5\)</p>

119 <p>= \(\frac{\frac{15}{80}}{\frac{30}{80}} = \frac{15}{30} = 0.5\)</p>

121 <p>Therefore, the probability of a pupil who learns English and Math at the same time is 0.5 or 50%.</p>

120 <p>Therefore, the probability of a pupil who learns English and Math at the same time is 0.5 or 50%.</p>

122 <p>Well explained 👍</p>

121 <p>Well explained 👍</p>

123 <h3>Problem 3</h3>

122 <h3>Problem 3</h3>

124 <p>On the turf, 70 children play badminton, and 30 play both badminton and basketball. If a child plays badminton, what is the probability that they also play basketball?</p>

123 <p>On the turf, 70 children play badminton, and 30 play both badminton and basketball. If a child plays badminton, what is the probability that they also play basketball?</p>

125 <p>Okay, lets begin</p>

124 <p>Okay, lets begin</p>

126 <p>The probability of a child playing both badminton and basketball is 42.86% or 0.43.</p>

125 <p>The probability of a child playing both badminton and basketball is 42.86% or 0.43.</p>

127 <h3>Explanation</h3>

126 <h3>Explanation</h3>

128 <p>We use the formula: </p>

127 <p>We use the formula: </p>

129 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

128 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

130 <p>Here, we have:</p>

129 <p>Here, we have:</p>

131 <p>P(A ∩ B) = 30 (children playing both badminton and basketball)</p>

130 <p>P(A ∩ B) = 30 (children playing both badminton and basketball)</p>

132 <p>P(B) = 70 (children playing badminton)</p>

131 <p>P(B) = 70 (children playing badminton)</p>

133 <p>Substituting the values:</p>

132 <p>Substituting the values:</p>

134 <p>\(P(A \mid B) = \frac{30}{70} = 0.4286\)</p>

133 <p>\(P(A \mid B) = \frac{30}{70} = 0.4286\)</p>

135 <p>Therefore, the probability of a child playing both badminton and basketball is 42.86% or 0.43.</p>

134 <p>Therefore, the probability of a child playing both badminton and basketball is 42.86% or 0.43.</p>

136 <p>Well explained 👍</p>

135 <p>Well explained 👍</p>

137 <h3>Problem 4</h3>

136 <h3>Problem 4</h3>

138 <p>A company manufactures 2,000 items daily. 600 come from Machine A, and 1400 from Machine B. Machine A makes 60 defective items, while Machine B makes 20 defective items. If an item is defective, what is the chance it came from Machine A?</p>

137 <p>A company manufactures 2,000 items daily. 600 come from Machine A, and 1400 from Machine B. Machine A makes 60 defective items, while Machine B makes 20 defective items. If an item is defective, what is the chance it came from Machine A?</p>

139 <p>Okay, lets begin</p>

138 <p>Okay, lets begin</p>

140 <p>The probability of a defective item from Machine A is 0.75 or 75%.</p>

139 <p>The probability of a defective item from Machine A is 0.75 or 75%.</p>

141 <h3>Explanation</h3>

140 <h3>Explanation</h3>

142 <p>We use the formula: </p>

141 <p>We use the formula: </p>

143 <p>\(P(A \mid D) = \frac{P(A \cap D)}{P(D)}\)</p>

142 <p>\(P(A \mid D) = \frac{P(A \cap D)}{P(D)}\)</p>

144 <p>P(A) = Probability of an item from Machine A = \(\frac{600}{2000} = 0.3\)</p>

143 <p>P(A) = Probability of an item from Machine A = \(\frac{600}{2000} = 0.3\)</p>

145 <p>P(B) = Probability of an item from Machine B = \(\frac{1400}{2000} = 0.7\)</p>

144 <p>P(B) = Probability of an item from Machine B = \(\frac{1400}{2000} = 0.7\)</p>

146 <p>P(D∣A) = Probability of a defective item from Machine A = \(\frac{60}{600} = 0.1\)</p>

145 <p>P(D∣A) = Probability of a defective item from Machine A = \(\frac{60}{600} = 0.1\)</p>

147 <p>P(D∣B) = Probability of a defective item from Machine B = \(\frac{20}{1400} = 0.0143\)</p>

146 <p>P(D∣B) = Probability of a defective item from Machine B = \(\frac{20}{1400} = 0.0143\)</p>

148 <p>Here, we use the total probability formula:</p>

147 <p>Here, we use the total probability formula:</p>

149 <p>\(P(D) = P(D|A) P(A) + P(D|B) P(B)\)</p>

148 <p>\(P(D) = P(D|A) P(A) + P(D|B) P(B)\)</p>

150 <p>\(P(D) = (0.1 × 0.3) + (0.0134 × 0.7)\)</p>

149 <p>\(P(D) = (0.1 × 0.3) + (0.0134 × 0.7)\)</p>

151 <p>\(= 0.03 + 0.01001\)</p>

150 <p>\(= 0.03 + 0.01001\)</p>

152 <p>\(= 0.04001\)</p>

151 <p>\(= 0.04001\)</p>

153 <p>\(≈ 0.04\)</p>

152 <p>\(≈ 0.04\)</p>

154 <p>Here, we apply Bayes’ theorem:</p>

153 <p>Here, we apply Bayes’ theorem:</p>

155 <p>\(P(A \mid D) = \frac{P(D \mid A)\, P(A)}{P(D)}\)</p>

154 <p>\(P(A \mid D) = \frac{P(D \mid A)\, P(A)}{P(D)}\)</p>

156 <p>\(= \frac{0.1 \times 0.3}{0.04} = \frac{0.03}{0.04} = 0.75\)</p>

155 <p>\(= \frac{0.1 \times 0.3}{0.04} = \frac{0.03}{0.04} = 0.75\)</p>

157 <p>Therefore, the probability of a defective item from machine A is 0.75 or 75%. </p>

156 <p>Therefore, the probability of a defective item from machine A is 0.75 or 75%. </p>

158 <p>Well explained 👍</p>

157 <p>Well explained 👍</p>

159 <h3>Problem 5</h3>

158 <h3>Problem 5</h3>

160 <p>A city records that 60% of youngsters drive above the speed limit. Among them, 25% are caught violating traffic rules. If a youngster is found violating traffic rules, what is the probability that they were also speeding?</p>

159 <p>A city records that 60% of youngsters drive above the speed limit. Among them, 25% are caught violating traffic rules. If a youngster is found violating traffic rules, what is the probability that they were also speeding?</p>

161 <p>Okay, lets begin</p>

160 <p>Okay, lets begin</p>

162 <p>0.75 is the probability.</p>

161 <p>0.75 is the probability.</p>

163 <h3>Explanation</h3>

162 <h3>Explanation</h3>

164 <p>Use the formula:</p>

163 <p>Use the formula:</p>

165 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

164 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

166 <p>A: The youngster was speeding</p>

165 <p>A: The youngster was speeding</p>

167 <p>B: The youngster was caught</p>

166 <p>B: The youngster was caught</p>

168 <p>\(P(A) = 0.60\)</p>

167 <p>\(P(A) = 0.60\)</p>

169 <p>\(P(B|A) = 0.25\)</p>

168 <p>\(P(B|A) = 0.25\)</p>

170 <p>\(P(B) = 0.20\)</p>

169 <p>\(P(B) = 0.20\)</p>

171 <p>\(P (A ∩ B) = P(B|A) P(A) = 0.25 × 0.60 = 0.15\)</p>

170 <p>\(P (A ∩ B) = P(B|A) P(A) = 0.25 × 0.60 = 0.15\)</p>

172 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

171 <p>\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)</p>

173 <p>\(= \frac{0.15}{0.20} = 0.75\)</p>

172 <p>\(= \frac{0.15}{0.20} = 0.75\)</p>

174 <p>Therefore, if a youngster violates the traffic rules, the probability that they were speeding is 75%.</p>

173 <p>Therefore, if a youngster violates the traffic rules, the probability that they were speeding is 75%.</p>

175 <p>Well explained 👍</p>

174 <p>Well explained 👍</p>

176 <h2>FAQs on Conditional Probability</h2>

175 <h2>FAQs on Conditional Probability</h2>

177 <h3>1.What is the formula for conditional probability?</h3>

176 <h3>1.What is the formula for conditional probability?</h3>

178 <p>Conditional probability can be mathematically expressed as A given B or P(A|B). </p>

177 <p>Conditional probability can be mathematically expressed as A given B or P(A|B). </p>

179 <p>P(A|B) = P(A ∩ B) / P(B)</p>

178 <p>P(A|B) = P(A ∩ B) / P(B)</p>

180 <p>Here:</p>

179 <p>Here:</p>

181 <p>P(A|B): The probability of occurrence of A is dependent on the likelihood of B occurring. </p>

180 <p>P(A|B): The probability of occurrence of A is dependent on the likelihood of B occurring. </p>

182 <p>P(A ∩ B): Probability of both events A and B occurring at the same time.</p>

181 <p>P(A ∩ B): Probability of both events A and B occurring at the same time.</p>

183 <p>P(B): Probability of the occurrence of event B.</p>

182 <p>P(B): Probability of the occurrence of event B.</p>

184 <h3>2.What is the conditional probability when P(B) equals 0?</h3>

183 <h3>2.What is the conditional probability when P(B) equals 0?</h3>

185 <p>Since<a>division by zero</a>is impossible, the conditional probability is undefined when P(B) = 0. </p>

184 <p>Since<a>division by zero</a>is impossible, the conditional probability is undefined when P(B) = 0. </p>

186 <p>If P(B) = 0, the conditional probability P(A|B) is undefined because division by zero is not possible. This means that event B will never occur, making it impossible to determine P(A|B).</p>

185 <p>If P(B) = 0, the conditional probability P(A|B) is undefined because division by zero is not possible. This means that event B will never occur, making it impossible to determine P(A|B).</p>

187 <h3>3.Cite one real-life example of conditional probability.</h3>

186 <h3>3.Cite one real-life example of conditional probability.</h3>

188 <p>Conditional probability predicts the likelihood of a person not going on a trip if it rains.</p>

187 <p>Conditional probability predicts the likelihood of a person not going on a trip if it rains.</p>

189 <h3>4.Is the value of conditional probability greater than 1?</h3>

188 <h3>4.Is the value of conditional probability greater than 1?</h3>

190 <p>No, the value of conditional probability is always between 0 and 1. </p>

189 <p>No, the value of conditional probability is always between 0 and 1. </p>

191 <h3>5.What is the major distinction between conditional probability and joint probability?</h3>

190 <h3>5.What is the major distinction between conditional probability and joint probability?</h3>

192 <p>Conditional probability predicts the probability of the occurrence of an event based on an event that has already occurred. However, joint probability predicts the probability of two events occurring simultaneously.</p>

191 <p>Conditional probability predicts the probability of the occurrence of an event based on an event that has already occurred. However, joint probability predicts the probability of two events occurring simultaneously.</p>

193 <h3>6.How can parents explain to their child what a conditional probability is?</h3>

192 <h3>6.How can parents explain to their child what a conditional probability is?</h3>

194 <p>Conditional probability is the chance that something will happen after another event has occurred. </p>

193 <p>Conditional probability is the chance that something will happen after another event has occurred. </p>

195 <h3>7.How can parents use daily situations to help their child understand conditional probability?</h3>

194 <h3>7.How can parents use daily situations to help their child understand conditional probability?</h3>

196 <p>Parents can use simple examples like: If your child has already finished homework, what is the chance they will get screen time? If it is cloudy, what is the chance your child will take an umbrella? </p>

195 <p>Parents can use simple examples like: If your child has already finished homework, what is the chance they will get screen time? If it is cloudy, what is the chance your child will take an umbrella? </p>

197 <h3>8.How can parents identify if their child is struggling with conditional probability?</h3>

196 <h3>8.How can parents identify if their child is struggling with conditional probability?</h3>

198 <p>If the child gets confused by the phrase “given that” or mixes up events, they may need extra help. Parents can use visual tools, such as Venn or tree diagrams, to clarify concepts.</p>

197 <p>If the child gets confused by the phrase “given that” or mixes up events, they may need extra help. Parents can use visual tools, such as Venn or tree diagrams, to clarify concepts.</p>

199 <h2>Jaipreet Kour Wazir</h2>

198 <h2>Jaipreet Kour Wazir</h2>

200 <h3>About the Author</h3>

199 <h3>About the Author</h3>

201 <p>Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref</p>

200 <p>Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref</p>

202 <h3>Fun Fact</h3>

201 <h3>Fun Fact</h3>

203 <p>: She compares datasets to puzzle games-the more you play with them, the clearer the picture becomes!</p>

202 <p>: She compares datasets to puzzle games-the more you play with them, the clearer the picture becomes!</p>