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3 Probability is the possibility that an event will occur. A discrete probability distribution describes the likelihood of outcomes for individually countable variables. In this topic, we will learn more about discrete probability distribution, its types, application, and many more.

4 <h2>What is a Discrete Probability Distribution?</h2>

5 The discrete<a>probability distribution</a>is a simple way to show the different chances<a>of</a>different outcomes when you can count them one by one. It gives the different values of a<a>random variable</a>along with its different probabilities. A discrete probability distribution contrasts with a continuous distribution, where outcomes can take any value along a continuum, as the outcome falls anywhere on a continuum. The types of discrete probability are Binomial, Poisson, and Bernoulli distributions. If a probability distribution is said to be a discrete probability distribution, it should follow these two conditions :

6 <ul><li>0 ≤ P(X = x) ≤ 1, which means the<a>discrete random variable</a>, X should be on the exact value, x, and it should be between 0 and 1</li>

7 <li>ΣP (X = x) = 1, that is, the<a>sum</a>of all probabilities should be 1. </li>

8 </ul><h3>Example:</h3>

9 Imagine you flip a fair coin 3 times, and you want to count the Number of Heads (X).

10 This is a perfect example because:

11 <ul><li>It is discrete:You can only get 0, 1, 2, or 3 heads. You can't get 1.5 heads.</li>

12 <li>It is finite:There is a limited<a>number</a>of possible outcomes.</li>

13 </ul>Step 1:List all possible outcomes

14 There are 8 total ways the coins can land ($2^3 = 8$):

15 <ul><li>3 Heads:HHH</li>

16 <li>2 Heads:HHT, HTH, THH</li>

17 <li>1 Head:HTT, THT, TTH</li>

18 <li>0 Heads:TTT.</li>

19 </ul>Step 2:Create the Distribution Table

20 Now, we count the probability for each number of heads.

21 Number of Heads (x)Frequency (How many ways?)Probability P(X=x)0 1 (TTT only) 1/8 (or 0.125) 1 3 (HTT, THT, TTH) 3/8 (or 0.375) 2 3 (HHT, HTH, THH) 3/8 (or 0.375) 3 1 (HHH only) 1/8 (or 0.125)<h2>Discrete Probability Distribution Formula</h2>

22 Here are the<a>formulas</a>for the PMF and CDF of Discrete Probability Distribution.

23 <h3>PMF of Discrete Probability Distribution</h3>

24 The PMF (Probability Mass Function) gives the<a>probability</a>that a discrete random<a>variable</a>X is exactly equal to a specific value x.

25 Formula:$f(x) = P(X = x)$

26 Example:Let X be the number of Girls in a family with 3 children.

27 Possible Outcomes ($2^3 = 8$ total outcomes):

28 <ul><li>GGG $\rightarrow$ 3 Girls</li>

29 <li>GGB, GBG, BGG $\rightarrow$ 2 Girls</li>

30 <li>GBB, BGB, BBG $\rightarrow$ 1 Girl</li>

31 <li>BBB $\rightarrow$ 0 Girls </li>

32 </ul>The PMFs are given as:

33 <ul><li>$P(X=0) = \frac{1}{8} \rightarrow$ (Only BBB)</li>

34 <li>$P(X=1) = \frac{3}{8} \rightarrow$ (GBB, BGB, BBG)</li>

35 <li>$P(X=2) = \frac{3}{8} \rightarrow$ (GGB, GBG, BGG)</li>

36 <li>$P(X=3) = \frac{1}{8} \rightarrow$ (Only GGG) </li>

37 </ul>Total Probability: $\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$

38 Properties of PMF:

39 <ol><li>The sum of all probabilities must equal 1: $\sum P(X=x) = 1$</li>

40 <li>Probabilities cannot be negative: $P(X=x) \ge 0$</li>

41 </ol><h3>CDF of Discrete Probability Distribution</h3>

42 The CDF (Cumulative Distribution Function) gives the probability that the random variable X is<a>less than</a>or equal to a specific value x.

43 Formula:$F(x) = P(X \le x)$

44 Example:Let a random variable X be the score obtained on a biased (weighted) 4-sided die. We want to find the probability $P(1 < X \le 3).$

45 Data Table:

46 X (Score)

47 PMF P(X=x)

48 CDF F(x)=P(X≤x)

49 1 0.1 0.1 2 0.4 0.1 + 0.4 = 0.5 3 0.3 0.5 + 0.3 = 0.8 4 0.2 0.8 + 0.2 = 1.0Calculation for $P(1 < X \le 3)$:

50 Using the CDF property: $P(a < X \le b) = F(b) - F(a)$

51 Therefore: $P(1 < X \le 3) = F(3) - F(1)$

52 $P(1 < X \le 3) = 0.8 - 0.1 = 0.7$

53 (Verification using PMF): $P(1 < X \le 3)$ means we want outcomes 2 and 3. $P(X=2) + P(X=3) = 0.4 + 0.3 = 0.7$

54 <h2>What are the Types of Discrete Probability Distributions</h2>

55 Now let’s discuss the types of discrete probability. The types are;

56 <ul><li>Binomial DistributionBinomial probability is where there is a probability of getting two outcomes. Here after repetitive trails the<a>data</a>is collected in one of the two forms, into either success or failure. For instance, when flipping a coin, the probability of getting heads. </li>

57 <li>Bernoulli DistributionIt is similar to<a>binomial</a>probability, as it also has two possible outcomes, with only one trial. Even here, the possible outcomes will be success or failure. </li>

58 <li>Multinomial DistributionIn multinomial distribution, the probability of getting more than two outcomes with<a>multiple</a>counts. For example, a bowl filled with three different colored coins like red, blue, and green. </li>

59 <li>Poisson DistributionIt is a type of probability that tells how a certain number of events happen in a fixed amount of time. For example, a Poisson distribution can be used for counting how many times a bus arrives at a stop in a fixed amount of time. </li>

60 <li>Geometric DistributionIt is a type of probability that tells you how many independent trials are needed to achieve the very first success. For example, a Geometric distribution can be used for counting how many job applications you need to submit until you finally get your first offer. </li>

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63 <h2>Mean of Discrete Probability Distribution</h2>

62 <h2>Mean of Discrete Probability Distribution</h2>

64 The Mean (also called the Expected Value) of a discrete probability distribution is the<a>weighted average</a>of all possible values that the random variable can take. It represents the long-<a>term</a>average result if you were to repeat the experiment many times.

63 The Mean (also called the Expected Value) of a discrete probability distribution is the<a>weighted average</a>of all possible values that the random variable can take. It represents the long-<a>term</a>average result if you were to repeat the experiment many times.

65 Formula:The<a>mean</a>\mu (or E(X)) is calculated by multiplying each possible outcome x by its probability P(X=x) and then adding them all together:

64 Formula:The<a>mean</a>\mu (or E(X)) is calculated by multiplying each possible outcome x by its probability P(X=x) and then adding them all together:

66 $\mu = E(X) = \sum [x \cdot P(X=x)]$

65 $\mu = E(X) = \sum [x \cdot P(X=x)]$

67 Example:Let a random variable X be the cash prize won in a charity raffle game.

66 Example:Let a random variable X be the cash prize won in a charity raffle game.

68 Possible Outcomes:

67 Possible Outcomes:

69 <ul><li>$0 (No win)</li>

68 <ul><li>$0 (No win)</li>

70 <li>$10 (Small prize)</li>

69 <li>$10 (Small prize)</li>

71 <li>$50 (Grand prize)</li>

70 <li>$50 (Grand prize)</li>

72 </ul>Calculation Table:We multiply each outcome (x) by its probability (P(x)) to find the “weighted” value.

71 </ul>Calculation Table:We multiply each outcome (x) by its probability (P(x)) to find the “weighted” value.

73 Outcome (x)

72 Outcome (x)

74 Probability P(X=x)

73 Probability P(X=x)

75 Weighted Value [x⋅P(x)]

74 Weighted Value [x⋅P(x)]

76 0 0.80 $0 \cdot 0.80 = \mathbf{0}$ 10 0.15 $10 \cdot 0.15 = \mathbf{1.5}$ 50 0.05 $50 \cdot 0.05 = \mathbf{2.5}$Total Mean:$E(X) = 0 + 1.5 + 2.5 = 4.0$

75 0 0.80 $0 \cdot 0.80 = \mathbf{0}$ 10 0.15 $10 \cdot 0.15 = \mathbf{1.5}$ 50 0.05 $50 \cdot 0.05 = \mathbf{2.5}$Total Mean:$E(X) = 0 + 1.5 + 2.5 = 4.0$

77 Interpretation:The expected value is $4. This means that if you played this game thousands of times, you would win an average of $4 per game, even though you can never actually win exactly $4 in a single turn.

76 Interpretation:The expected value is $4. This means that if you played this game thousands of times, you would win an average of $4 per game, even though you can never actually win exactly $4 in a single turn.

78 <h2>Variance of Discrete Probability Distribution</h2>

77 <h2>Variance of Discrete Probability Distribution</h2>

79 The Variance measures how “spread out” or dispersed the numbers are from the mean (expected value). A high<a>variance</a>means the values are widely scattered; a low variance means they are clustered closely around the<a>average</a>.

78 The Variance measures how “spread out” or dispersed the numbers are from the mean (expected value). A high<a>variance</a>means the values are widely scattered; a low variance means they are clustered closely around the<a>average</a>.

80 Formula:There are two common ways to calculate it.

79 Formula:There are two common ways to calculate it.

81 <ol><li>The Definition Formula: The weighted average of the squared differences from the mean.$\sigma^2 = Var(X) = \sum [(x - \mu)^2 \cdot P(x)]$

80 <ol><li>The Definition Formula: The weighted average of the squared differences from the mean.$\sigma^2 = Var(X) = \sum [(x - \mu)^2 \cdot P(x)]$

82 </li>

81 </li>

83 <li>The Computational Formula (Often Easier): The average of the<a>squares</a>minus the square of the average.$\sigma^2 = E(X^2) - [E(X)]^2 = \sum [x^2 \cdot P(x)] - \mu^2$

82 <li>The Computational Formula (Often Easier): The average of the<a>squares</a>minus the square of the average.$\sigma^2 = E(X^2) - [E(X)]^2 = \sum [x^2 \cdot P(x)] - \mu^2$

84 </li>

83 </li>

85 </ol>Example:Let a random variable X be the number of daily complaints received by a service center.

84 </ol>Example:Let a random variable X be the number of daily complaints received by a service center.

86 Possible Outcomes:

85 Possible Outcomes:

87 <ul><li>0 complaints (Quiet day)</li>

86 <ul><li>0 complaints (Quiet day)</li>

88 <li>1 complaint (Normal day)</li>

87 <li>1 complaint (Normal day)</li>

89 <li>2 complaints (Busy day)</li>

88 <li>2 complaints (Busy day)</li>

90 </ul>Probabilities:

89 </ul>Probabilities:

91 <ul><li>P(X=0) = 0.1</li>

90 <ul><li>P(X=0) = 0.1</li>

92 <li>P(X=1) = 0.6</li>

91 <li>P(X=1) = 0.6</li>

93 <li>P(X=2) = 0.3</li>

92 <li>P(X=2) = 0.3</li>

94 </ul>Step 1:Find the Mean (\mu) $\mu = (0 \cdot 0.1) + (1 \cdot 0.6) + (2 \cdot 0.3)$ $\mu = 0 + 0.6 + 0.6 = \mathbf{1.2}$ $\mu^2 = 1.2^2 = 1.44$

93 </ul>Step 1:Find the Mean (\mu) $\mu = (0 \cdot 0.1) + (1 \cdot 0.6) + (2 \cdot 0.3)$ $\mu = 0 + 0.6 + 0.6 = \mathbf{1.2}$ $\mu^2 = 1.2^2 = 1.44$

95 Step 2:Calculate $E(X^2)$ We need to square each outcome first, then multiply by its probability

94 Step 2:Calculate $E(X^2)$ We need to square each outcome first, then multiply by its probability

96 Outcome (x)

95 Outcome (x)

97 Probability P(x)

96 Probability P(x)

98 Squared Outcome $(x^2)$Weighted Square$E(X^2) = [x^2⋅P(x)]$0 0.1 $0^2 = 0$ $0 \cdot 0.1 = \mathbf{0}$ 1 0.6 $1^2 = 1$ $1 \cdot 0.6 = \mathbf{0.6}$ 2 0.3 $2^2 = 4$ $4 \cdot 0.3 = \mathbf{1.2}$$\sum$ $0+0.6+1.2=1.8$Step 3:Apply the Formula Now we subtract the square of the mean from the sum we just found.

97 Squared Outcome $(x^2)$Weighted Square$E(X^2) = [x^2⋅P(x)]$0 0.1 $0^2 = 0$ $0 \cdot 0.1 = \mathbf{0}$ 1 0.6 $1^2 = 1$ $1 \cdot 0.6 = \mathbf{0.6}$ 2 0.3 $2^2 = 4$ $4 \cdot 0.3 = \mathbf{1.2}$$\sum$ $0+0.6+1.2=1.8$Step 3:Apply the Formula Now we subtract the square of the mean from the sum we just found.

99 Variance($\sigma^2$) $\sigma^2 = E(X^2) - \mu^2$ $\sigma^2 = 1.8 - 1.44$ $\sigma^2 = \mathbf{0.36}$

98 Variance($\sigma^2$) $\sigma^2 = E(X^2) - \mu^2$ $\sigma^2 = 1.8 - 1.44$ $\sigma^2 = \mathbf{0.36}$

100 Standard deviation($\sigma$)$\sigma = \sqrt{0.36} = \mathbf{0.6}$

99 Standard deviation($\sigma$)$\sigma = \sqrt{0.36} = \mathbf{0.6}$

101 <h2>How to Calculate Discrete Probability Distribution</h2>

100 <h2>How to Calculate Discrete Probability Distribution</h2>

102 Here, we will discuss how to calculate a discrete probability distribution. To find a discrete probability distribution, the probability of mass<a>function</a>is required. Follow these steps to find the probability;

101 Here, we will discuss how to calculate a discrete probability distribution. To find a discrete probability distribution, the probability of mass<a>function</a>is required. Follow these steps to find the probability;

103 Step 1:Identify the sample,<a>set</a>of all possible outcomes of an experiment.

102 Step 1:Identify the sample,<a>set</a>of all possible outcomes of an experiment.

104 Step 2:Then you have to find the discrete random variable (x). It is a function assigning a numerical value to each outcome.

103 Step 2:Then you have to find the discrete random variable (x). It is a function assigning a numerical value to each outcome.

105 Step 3:Identifying the possible value of X.

104 Step 3:Identifying the possible value of X.

106 Step 4:Finding the probability of each outcome by using the formula;

105 Step 4:Finding the probability of each outcome by using the formula;

107 P (X = x) = Number of favorable outcomes/Total number of possible outcomes.

106 P (X = x) = Number of favorable outcomes/Total number of possible outcomes.

108 Step 5:Use a table to organize the results

107 Step 5:Use a table to organize the results

109 <h2>Discrete Distribution vs. Continuous Distribution</h2>

108 <h2>Discrete Distribution vs. Continuous Distribution</h2>

110 Now let’s learn the difference between the discrete and continuous distribution.

109 Now let’s learn the difference between the discrete and continuous distribution.

111 Discrete Distribution

110 Discrete Distribution

112 Continuous Distribution

111 Continuous Distribution

113 The probability distribution for countable values

112 The probability distribution for countable values

114 The probability distribution for measurable values

113 The probability distribution for measurable values

115 Here, the type of variable is discrete

114 Here, the type of variable is discrete

116 Here, the type of variable is continuous

115 Here, the type of variable is continuous

117 The graph of discrete distribution is a bar

116 The graph of discrete distribution is a bar

118 The graph of continuous distribution is a curve

117 The graph of continuous distribution is a curve

119 <h2>Tips and Tricks to Master Discrete Probability Distribution</h2>

118 <h2>Tips and Tricks to Master Discrete Probability Distribution</h2>

120 Discrete probability distribution is a complex topic to get a grasp on. In this section, we will discuss some tips and tricks to master Discrete probability distribution.

119 Discrete probability distribution is a complex topic to get a grasp on. In this section, we will discuss some tips and tricks to master Discrete probability distribution.

121 <ul><li>Organize your data in a simple table with two columns:One for the random variable (X) and one for its probability P(X). </li>

120 <ul><li>Organize your data in a simple table with two columns:One for the random variable (X) and one for its probability P(X). </li>

122 <li>Practice with real-life scenarios:Apply discrete probability to everyday examples like tossing coins, quality checks, or customer arrivals. </li>

121 <li>Practice with real-life scenarios:Apply discrete probability to everyday examples like tossing coins, quality checks, or customer arrivals. </li>

123 <li>Identify the type of distribution:Understand whether the problem is about a fixed number of trials, rare events, or waiting for the first success. Correctly identifying the type helps you approach it the right way. </li>

122 <li>Identify the type of distribution:Understand whether the problem is about a fixed number of trials, rare events, or waiting for the first success. Correctly identifying the type helps you approach it the right way. </li>

124 <li>Understand conditions:Check the assumptions behind each distribution, like independence of events or whether outcomes are countable. This prevents applying the wrong distribution. </li>

123 <li>Understand conditions:Check the assumptions behind each distribution, like independence of events or whether outcomes are countable. This prevents applying the wrong distribution. </li>

125 <li>Use expected patterns:Think about what results are reasonable based on the situation. This helps spot errors in calculations or assumptions.</li>

124 <li>Use expected patterns:Think about what results are reasonable based on the situation. This helps spot errors in calculations or assumptions.</li>

126 </ul><h2>Common mistakes and How to Avoid Them in Discrete Probability Distribution</h2>

125 </ul><h2>Common mistakes and How to Avoid Them in Discrete Probability Distribution</h2>

127 Now let’s learn a few common mistakes and ways to avoid them to master discrete probability distribution.

126 Now let’s learn a few common mistakes and ways to avoid them to master discrete probability distribution.

128 <h2>Real-Life Applications of Discrete Probability Distribution</h2>

127 <h2>Real-Life Applications of Discrete Probability Distribution</h2>

129 In real-life discrete probability distribution are used to find the probability for countable values, now lets few real-world applications of it;

128 In real-life discrete probability distribution are used to find the probability for countable values, now lets few real-world applications of it;

130 <ul><li>For quality control in manufacturing, we use<a>binomial distribution</a>to find the number of defective products in a batch based on the random checks. </li>

129 <ul><li>For quality control in manufacturing, we use<a>binomial distribution</a>to find the number of defective products in a batch based on the random checks. </li>

131 </ul><ul><li>When playing the games of chances, like for example cards, snake and ladder, etc. Probability helps you figure out how likely you are to win. </li>

130 </ul><ul><li>When playing the games of chances, like for example cards, snake and ladder, etc. Probability helps you figure out how likely you are to win. </li>

132 </ul><ul><li>The number of calls received in a call center is predicted using the Poisson distribution.</li>

131 </ul><ul><li>The number of calls received in a call center is predicted using the Poisson distribution.</li>

133 </ul><ul><li>To identify whether an email is a spam or not using the Bernoulli distribution. </li>

132 </ul><ul><li>To identify whether an email is a spam or not using the Bernoulli distribution. </li>

134 <li>Insurance companies use the normal distribution to predict risk and estimate claims. </li>

133 <li>Insurance companies use the normal distribution to predict risk and estimate claims. </li>

135 </ul><h3>Problem 1</h3>

134 </ul><h3>Problem 1</h3>

136 A factory produces light bulbs, and 5% of them are defective. If a random sample of 8 bulbs is taken, what is the probability that exactly 2 bulbs are defective?

135 A factory produces light bulbs, and 5% of them are defective. If a random sample of 8 bulbs is taken, what is the probability that exactly 2 bulbs are defective?

137 Okay, lets begin

136 Okay, lets begin

138 The probability of getting 2 bulbs are 0.0515 or 5.15%.

137 The probability of getting 2 bulbs are 0.0515 or 5.15%.

139 <h3>Explanation</h3>

138 <h3>Explanation</h3>

140 Here, to find the probability we use the equation, P(X = k) = $\binom{n}{k}$ pk(1 - p)n-k

139 Here, to find the probability we use the equation, P(X = k) = $\binom{n}{k}$ pk(1 - p)n-k

141 According to the problem,

140 According to the problem,

142 n = 8

141 n = 8

143 k = 2

142 k = 2

144 p = 0.05

143 p = 0.05

145 Substituting the values,

144 Substituting the values,

146 P(X = 2) = $\binom{8}{2}$ (0.05)2(1 - 0.05)8-2

145 P(X = 2) = $\binom{8}{2}$ (0.05)2(1 - 0.05)8-2

147 P(X = 2) = $\binom{8}{2}$ (0.05)2( 0.95)6

146 P(X = 2) = $\binom{8}{2}$ (0.05)2( 0.95)6

148 The value of C $\binom{8}{2}$= 28, (0.05)2 = 0.0025, and (0.95)6 = 0.7358

147 The value of C $\binom{8}{2}$= 28, (0.05)2 = 0.0025, and (0.95)6 = 0.7358

149 So, the value of P(X = 2) = 28 × 0.0025 × 0.7358 = 0.0515.

148 So, the value of P(X = 2) = 28 × 0.0025 × 0.7358 = 0.0515.

150 Well explained 👍

149 Well explained 👍

151 <h3>Problem 2</h3>

150 <h3>Problem 2</h3>

152 A basketball player has a 40% probability of making a free throw. What is the probability that their first successful shot happens on the third attempt?

151 A basketball player has a 40% probability of making a free throw. What is the probability that their first successful shot happens on the third attempt?

153 Okay, lets begin

152 Okay, lets begin

154 The probability here is 0.144 or 14.4%.

153 The probability here is 0.144 or 14.4%.

155 <h3>Explanation</h3>

154 <h3>Explanation</h3>

156 Here the probability is of geometric distribution

155 Here the probability is of geometric distribution

157 So, P(X = k) = (1 - p)(k-1) × p.

156 So, P(X = k) = (1 - p)(k-1) × p.

158 Here, p = 0.4 and k = 3

157 Here, p = 0.4 and k = 3

159 Therefore, P(X = 3) = (0.6)2 × 0.4

158 Therefore, P(X = 3) = (0.6)2 × 0.4

160 = 0.36 × 0.4 = 0.144.

159 = 0.36 × 0.4 = 0.144.

161 Well explained 👍

160 Well explained 👍

162 <h3>Problem 3</h3>

161 <h3>Problem 3</h3>

163 A deck has 52 cards, including 4 Aces. If 5 cards are drawn randomly, what is the probability of getting exactly 2 Aces?

162 A deck has 52 cards, including 4 Aces. If 5 cards are drawn randomly, what is the probability of getting exactly 2 Aces?

164 Okay, lets begin

163 Okay, lets begin

165 Probability ≈ 0.0399 or 3.99%.

164 Probability ≈ 0.0399 or 3.99%.

166 <h3>Explanation</h3>

165 <h3>Explanation</h3>

167 Here, the probability is calculated using $P (X = k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}.$

166 Here, the probability is calculated using $P (X = k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}.$

168 Here,

167 Here,

169 N = 52

168 N = 52

170 K = 4

169 K = 4

171 k = 2

170 k = 2

172 n = 5

171 n = 5

173 So P(X = 2) =$ \dfrac{\binom{4}{2} \binom{52 - 4}{5 - 2}}{\binom{52}{5}}.$

172 So P(X = 2) =$ \dfrac{\binom{4}{2} \binom{52 - 4}{5 - 2}}{\binom{52}{5}}.$

174 P(X = 2) = 0.0399 or 3.99%.

173 P(X = 2) = 0.0399 or 3.99%.

175 Well explained 👍

174 Well explained 👍

176 <h3>Problem 4</h3>

175 <h3>Problem 4</h3>

177 A multiple-choice quiz has 10 questions, each with 4 answer choices. A student guesses randomly on each question. What is the probability of getting exactly 3 correct answers?

176 A multiple-choice quiz has 10 questions, each with 4 answer choices. A student guesses randomly on each question. What is the probability of getting exactly 3 correct answers?

178 Okay, lets begin

177 Okay, lets begin

179 The probability is 0.250 or 25%.

178 The probability is 0.250 or 25%.

180 <h3>Explanation</h3>

179 <h3>Explanation</h3>

181 Here, to find the probability we use the equation, P(X = k) = C $\binom{n}{k}$ pk (1 - p)(n - k)

180 Here, to find the probability we use the equation, P(X = k) = C $\binom{n}{k}$ pk (1 - p)(n - k)

182 According to the problem,

181 According to the problem,

183 n = 10

182 n = 10

184 k = 3

183 k = 3

185 p = ¼ = 0.25

184 p = ¼ = 0.25

186 Substituting the values,

185 Substituting the values,

187 P(X = 3) = $\binom{10}{3}$(0.25)2(1 - 0.25)10 - 3

186 P(X = 3) = $\binom{10}{3}$(0.25)2(1 - 0.25)10 - 3

188 P(X = 3) = $\binom{10}{3}$ (0.25)2(0.75)7

187 P(X = 3) = $\binom{10}{3}$ (0.25)2(0.75)7

189 Here, C$\binom{10}{3}$ = 120

188 Here, C$\binom{10}{3}$ = 120

190 0.252 = 0.015625

189 0.252 = 0.015625

191 0.757 = 0.1335

190 0.757 = 0.1335

192 P(X = 3) ≈ 120 × 0.015625 × 0.1335 ≈ 0.250.

191 P(X = 3) ≈ 120 × 0.015625 × 0.1335 ≈ 0.250.

193 Well explained 👍

192 Well explained 👍

194 <h3>Problem 5</h3>

193 <h3>Problem 5</h3>

195 A factory produces screws with a 2% defect rate. What is the probability that the first defective screw appears on the 5th inspection?

194 A factory produces screws with a 2% defect rate. What is the probability that the first defective screw appears on the 5th inspection?

196 Okay, lets begin

195 Okay, lets begin

197 The probability is 0.01845 or 1.845%.

196 The probability is 0.01845 or 1.845%.

198 <h3>Explanation</h3>

197 <h3>Explanation</h3>

199 P(X = 5) = (1 - p)(k-1) × p.

198 P(X = 5) = (1 - p)(k-1) × p.

200 Here, p = 0.02

199 Here, p = 0.02

201 K = 5

200 K = 5

202 P(X=5) = (1 - p)(k-1)p = (0.98)4 × 0.02

201 P(X=5) = (1 - p)(k-1)p = (0.98)4 × 0.02

203 As 0.984 = 0.9224

202 As 0.984 = 0.9224

204 P(X = 5) = 0.9224 × 0.02 = 0.01845.

203 P(X = 5) = 0.9224 × 0.02 = 0.01845.

205 Well explained 👍

204 Well explained 👍

206 <h2>FAQs on Discrete Probability Distribution</h2>

205 <h2>FAQs on Discrete Probability Distribution</h2>

207 <h3>1.What is a discrete probability distribution?</h3>

206 <h3>1.What is a discrete probability distribution?</h3>

208 A discrete probability distribution is one type of probability that shows different chances of different countable outcomes of an event that are to happen.

207 A discrete probability distribution is one type of probability that shows different chances of different countable outcomes of an event that are to happen.

209 <h3>2.What is the difference between discrete probability distribution and continuous probability?</h3>

208 <h3>2.What is the difference between discrete probability distribution and continuous probability?</h3>

210 Discrete probability distribution deals with counted numbers whereas continuous probability deals with the measurements

209 Discrete probability distribution deals with counted numbers whereas continuous probability deals with the measurements

211 <h3>3.What are the real-life applications of discrete probability distribution?</h3>

210 <h3>3.What are the real-life applications of discrete probability distribution?</h3>

212 Discrete probability distribution is used in real-life situations to find customers' arrivals, phone calls, chances of winning a game and so on.

211 Discrete probability distribution is used in real-life situations to find customers' arrivals, phone calls, chances of winning a game and so on.

213 <h3>4.What are the types of discrete probability distribution?</h3>

212 <h3>4.What are the types of discrete probability distribution?</h3>

214 Bernoulli distribution, binomial distribution, Poisson distribution, and multinomial distribution are the four types of probability distributions.

213 Bernoulli distribution, binomial distribution, Poisson distribution, and multinomial distribution are the four types of probability distributions.

215 <h3>5.What are the conditions that a discrete probability distribution follows?</h3>

214 <h3>5.What are the conditions that a discrete probability distribution follows?</h3>

216 Discrete probability follows two conditions, that is p(x) ≥ 0 and Σp(x) = 1

215 Discrete probability follows two conditions, that is p(x) ≥ 0 and Σp(x) = 1

217 <h2>Jaipreet Kour Wazir</h2>

216 <h2>Jaipreet Kour Wazir</h2>

218 <h3>About the Author</h3>

217 <h3>About the Author</h3>

219 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

218 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

220 <h3>Fun Fact</h3>

219 <h3>Fun Fact</h3>

221 : She compares datasets to puzzle games-the more you play with them, the clearer the picture becomes!

220 : She compares datasets to puzzle games-the more you play with them, the clearer the picture becomes!