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Chapter 27: Problem 65

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is [2004] (a) \(\frac{28}{256}\) (b) \(\frac{219}{256}\) (c) \(\frac{128}{256}\) (d) \(\frac{37}{256}\)

Short Answer

Expert verified
The probability of 2 successes is option (a) \( \frac{28}{256} \).

Step by step solution

01

Understanding the Binomial Distribution

A binomial distribution is defined by the parameters \( n \) (number of trials) and \( p \) (probability of success in each trial). The mean \( \mu \) of a binomial distribution is given by \( \mu = np \) and the variance \( \sigma^2 \) is given by \( \sigma^2 = np(1-p) \). We're given \( \mu = 4 \) and \( \sigma^2 = 2 \).
02

Establish Equations from Mean and Variance

From the mean equation, we have: \( np = 4 \). From the variance equation, we have: \( np(1-p) = 2 \). We will use these equations to find \( n \) and \( p \).
03

Solve for Probability \( p \)

Substitute \( np = 4 \) into the variance equation:\[ np(1-p) = 2 \Rightarrow 4(1-p) = 2 \]Solving gives:\[ 4 - 4p = 2 \Rightarrow 4p = 2 \Rightarrow p = \frac{1}{2} \]
04

Solve for Number of Trials \( n \)

Now that we have \( p = \frac{1}{2} \), substitute back into \( np = 4 \):\[ n \cdot \frac{1}{2} = 4 \Rightarrow n = 8 \]
05

Use Binomial Formula for Probability of 2 Successes

The probability of exactly \( k \) successes in \( n \) trials is given by:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]For 2 successes out of 8 trials with \( p = \frac{1}{2} \):\[ P(X = 2) = \binom{8}{2} \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^{8-2} = \binom{8}{2} \left( \frac{1}{2} \right)^8 \]
06

Compute the Binomial Coefficient and Probability

Compute the binomial coefficient:\[ \binom{8}{2} = \frac{8 imes 7}{2 imes 1} = 28 \]Therefore:\[ P(X = 2) = 28 \times \frac{1}{256} = \frac{28}{256} \]
07

Select the Correct Answer

The probability of 2 successes is \( \frac{28}{256} \). Match this with the options given, the answer is (a) \( \frac{28}{256} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean
In a binomial distribution, the mean is a measure of central tendency that shows what you can expect as an average number of successes in a series of trials. It's determined by the formula \( \mu = np \), where \( n \) is the number of trials and \( p \) is the probability of success on a single trial.

For example, if you flip a coin 8 times, and the probability of getting heads on a single flip is \( \frac{1}{2} \), then you can use the formula \( \mu = 8 \times \frac{1}{2} = 4 \). This indicates that, on average, 4 heads can be expected.

Understanding the mean helps you predict outcomes when facing uncertainty, like estimating the average number of patients visiting a clinic in a week based on past data.
Variance
Variance in a binomial distribution describes the spread of the distribution or how much the outcomes deviate from the mean. It's given by the formula \( \sigma^2 = np(1-p) \).

The term \( (1-p) \) represents the probability of failure in each trial, providing insight into not only how often successes occur but also when they don't. For instance, if the mean is 4 and variance is 2, this shows both a balanced number of successes and a notable spread of results.

Calculating variance is essential in predicting unexpected deviations. For example, a high variance could indicate a wider range of customer purchases, which could influence stocking decisions in a store.
Probability of Success
The probability of success in a binomial distribution is fundamental for predicting outcomes. It is denoted by \( p \) and represents the likelihood of achieving a success in a single trial.

For instance, when we say the probability of rolling a four with a die is \( \frac{1}{6} \), this is our \( p \) value. Understanding this concept allows for calculations of more complex probabilities, like the probability of rolling a four twice in three rolls.

In the given problem, once the mean and variance were found, determining \( p \) was crucial to finding the probability of exactly 2 successes in 8 trials using the binomial formula. Mastering these calculations allows for accurate predictions of outcomes in real-world scenarios, like determining the chance of delivering a package by a certain deadline.

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Most popular questions from this chapter

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is: [2013] (a) \(\frac{17}{3^{5}}\) (b) \(\frac{13}{3^{5}}\) (c) \(\frac{11}{3^{5}}\) (d) \(\frac{10}{3^{5}}\)

Let \(\mathrm{A}\) and \(\mathrm{B}\) be two non-null events such that \(\mathrm{A} \subset \mathrm{B}\). Then, which of the following statements is always correct? [April \(08,2019(\) I) \(]\) (a) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})\) (b) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \geq \mathrm{P}(\mathrm{A})\) (c) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \leq \mathrm{P}(\mathrm{A})\) (d) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B})=1\)

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are \(0.3\) and \(0.2\), respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is (a) \(0.2\) (b) \(0.7\) (c) \(0.06\) (d) \(0.14\)

If two different numbers are taken from the set \((0,1,2,3,\), \(\ldots \ldots, 10\) ), then the probability that their sum as well as absolute difference are both multiple of 4 , is: (a) \(\frac{7}{55}\) (b) \(\frac{6}{55}\) (c) \(\frac{12}{55}\) (d) \(\frac{14}{55}\)

A player \(X\) has a biased coin whose probability of showing heads is \(p\) and a player \(Y\) has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If \(X\) starts the game, and the probability of winning the game by both the players is equal, then the value of ' \(p\) ' is[Online April 15, 2018] (a) \(\frac{1}{3}\) (b) \(\frac{1}{5}\) (c) \(\frac{1}{4}\) (d) \(\frac{2}{5}\)
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