Question

A box contains \[100\] bulbs, out of which \[10\] are defective. A sample of \[5\] bulbs is drawn. The probability that none is defective, is
1. \[\dfrac{9}{10}\]
2. \[{{\left( \dfrac{1}{10} \right)}^{5}}\]
3. \[{{\left( \dfrac{9}{10} \right)}^{5}}\]
4. \[{{\left( \dfrac{1}{2} \right)}^{5}}\]

Answer 1

Hint: Now to solve this question here in this question we can find the probability of how many bulbs will be defective out of the total amount of bulbs and we will also be able to find the probability of how many bulbs will not be faulty. Then we can use the formula for binomial distribution of probability to find the probability that none of the bulbs are defective. The formula is
\[P(X=x){{=}^{n}}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\]

Complete step-by-step answer:
Now here in this question it is said that out of \[100\] bulbs there are \[10\] bulbs which are defective. Therefore using this information we can find that
Probability of the bulb drawn to be defective \[=\dfrac{10}{100}=\dfrac{1}{10}\]
Now since a bulb is either defective or either not defective we can say that the sum of their probability will be one. Therefore we can say that
Probability of the bulb drawn to not be defective \[=1-\dfrac{1}{10}=\dfrac{9}{10}\]
Now we draw \[5\]bulbs as said in the question so we can say that \[n=5\]
Thus we can say that \[X\]has a binomial distribution of \[n=5\] where
\[p=\dfrac{1}{10}\] because p is the probability of the bulb drawn being defective
\[q=\dfrac{9}{10}\] because q is the probability of the bulb drawn not being defective
Now using the formula of binomial distribution for probability
\[P(X=x){{=}^{n}}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\]
Substituting
\[P(X=x){{=}^{5}}{{C}_{x}}{{\left( \dfrac{1}{10} \right)}^{x}}{{\left( \dfrac{9}{10} \right)}^{5-x}}\]
Now to find the probability of none of the bulbs being defective is, we put x as zero.
\[P(X=0){{=}^{5}}{{C}_{0}}{{\left( \dfrac{1}{10} \right)}^{0}}{{\left( \dfrac{9}{10} \right)}^{5-0}}\]
Solving using the formula for combinations we get, also anything raised to zero is one.
\[P(X=0)=\dfrac{5!}{0!\left( 5-1 \right)!}\times 1\times {{\left( \dfrac{9}{10} \right)}^{5}}\]
Hence
\[P(X=0)=\dfrac{5!}{1\times 5!}\times 1\times {{\left( \dfrac{9}{10} \right)}^{5}}\]
Cancelling the factorial
\[P(X=0)=1\times 1\times {{\left( \dfrac{9}{10} \right)}^{5}}\]
So we get the answer that
\[P(X=0)={{\left( \dfrac{9}{10} \right)}^{5}}\]
So the probability of none of bulbs being defective is \[{{\left( \dfrac{9}{10} \right)}^{5}}\]
So, the correct answer is “Option 3”.

Note: Here a common mistake that can be made while solving is we could alternate the value of the probability of bulb being defective and not being defective which could give us the wrong answer so always make sure the value of p and q are right and you put the x accordingly as asked in the question to always get the correct answer.