Question

Ten percent of the tools produced by a company turn out to be faulty. Find the probability that in a sample of 10 tools chosen at random, exactly 2 would be faulty by using (I) the binomial distribution (ii) the poisson approximation to the binomial

Answer 1

ANSWER

Let the event of a tool being defective be X.

∴ Probability of X occurring = p=101

∴q=1−p=109 

Now, as the process follows the binomial distribution, the probability of 2 tools being defective is

10C2p2q10−2

=10C2p2q8

=2!8!10!(101)2(109)8

=0.194

This is the required solution.

Answer 2

Answer:

Binomial distribution is equal to $0.194.$

The Poisson approximation to the binomial is $\frac{1}{2e}$.

Step-by-step explanation:

Probability: The concept of a random event is covered in the mathematical field of probability. For instance, if a coin is flung into the air, it could land on its Head or Tail.

Probability= number of favorable outcomes/ Total number of outcomes

Binomial Distribution: When each trial has the same probability of achieving a given value, the number of trials or observations is summarized using the binomial distribution. The likelihood of observing a specific number of successful outcomes in a specific number of trials is determined by the binomial distribution.

The Poisson approximation to the binomial: The Poisson distribution is actually a limiting case of a Binomial distribution when the number of trials, n, gets very large and p, the probability of success, is small. As a rule of thumb, if n≥100 and np≤10, the Poisson distribution (taking λ=np) can provide a very good approximation to the binomial distribution.

Given:

Let the event of tools being defective be $Y.$

So, probability of $Y$ occurring$=P(Y)=\frac{1}{10}$

Probability of occurring non defective tools

$=Q(Y)=1-P(Y)$

$=1-\frac{1}{10} \\=\frac{9}{10}$

(i) Now, as the process follows the binomial distribution, the probability of 2 tools being defective is:

$C_{2}^{10}p^{2}q^{^{10-2}}$

$=C_{2}^{10}p^{2}q^{^{10-2}}$

$=\frac{10!}{2!8!} \left ( \frac{1}{10} \right )^{^{2}}\left ( \frac{9}{10} \right )^{8}\\=0.194$

(ii)  The Poisson approximation to the binomial

$P(Y=y)=\frac{e^{-\lambda }\lambda ^{y}}{y!}$

where, $\lambda =$mean$=n\cdot P(Y)$

n is total number of tools produced found to be faulty

$n=10$

$\lambda =10\times \frac{1}{10}\\\lambda =1$

So,

$P(Y=2)=\frac{e^{-1}(1)^{2}}{2!}$

$=\frac{e^{-1}}{2!}\\\\=\frac{1}{2e}$

Hence binomial distribution is equal to $0.194.$

The Poisson approximation to the binomial is $\frac{1}{2e}$.

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