Question

A box contains 10 items, 3 of which are defective. If 4 are selected at random without replacement, the probability that at least 2 of them are defective is?
A.50%
B.33.33%
C.67%
D.100%

Answer 1

Hint: In this question, the total number of items is 10 and the number of defective items is 3. We are going to have take cases for the same (because it has been given to evaluate for at least 2 items to be defective) – if 2 out of 4 are defective or if 3 out 4 are defective, and it cannot be for if 1 out of 4 is defective as at least 2 items have to be defective and nor can it be if 4 out of 4 are defective as only 3 items are defective. So, we only have two cases, and we have to apply the required formula and calculate the required probability.
Formula Used:
Let the favorable probability of an event be P(E) with the total number of items being n and the number of favorable items be m.
Then, the value of P(E) (in percent) is given by:
 \[P\left( E \right) = \dfrac{m}{n} \times 100\]

Complete step-by-step answer:
In this question, we are going to consider 2 cases (because it has been given that there should be at least 2 defective items)
Total number of defective items = 3
If 2 of the randomly picked items are defective,
\[{P_1}{ = ^3}{C_2}{ \times ^7}{C_2} = \dfrac{{3!}}{{2!}} \times \dfrac{{7!}}{{2! \times 5!}} = 3 \times \dfrac{{7 \times 6}}{2} = 63\]
If 3 of the randomly picked items are defective,
\[{P_2}{ = ^3}{C_3}{ \times ^7}{C_1} = 1 \times 7 = 7\]
So, total number of favorable events,
\[P = 7 + 63 = 70\]
Now, the total number of events given that there are 4 trials,
\[N{ = ^{10}}{C_4} = \dfrac{{10!}}{{6! \times 4!}} = \dfrac{{10 \times 9 \times 8 \times 7}}{{24}} = 210\]
So, the required probability
\[P\left( E \right) = \dfrac{P}{N} = \dfrac{{70}}{{210}} = \dfrac{1}{3}\]
In terms of percent, \[P\left( E \right) = \dfrac{1}{3} \times 100 = 33.33\% \]
So, the correct answer is “Option B”.

Note: So, we saw that in solving questions like these, we need to consider all the possible cases of the possibility of the required event. But, we need to check if any occurring case is out of bounds, like here for the 4 out of 4 defective item case, it is not possible as it is out of bounds because there are a maximum of 3 defective items in there. Then, after separately evaluating all the possible cases of the required event, we need to add the individual probability to finally arrive at the final answer.