Question

If a box contains 10 items, out of which 3 items are found to be defective. Assume if 4 items are drawn at random without replacement, what is the probability that at least 2 out of 4 are found to be defective?
(a) 0.009
(b) 0.047
(c) 0.026
(d) None of these

Answer 1

Hint : We first calculate the probability such that 2 out of 4 drawn are defective if 4 items are drawn at random without replacement. Then we calculate the probability such that 3 out of 4 drawn are defective if 4 items are drawn at random without replacement. Since, we need to find the probability that at least 2 out of 4 are found to be defective we add both obtained values to the required result.

Complete step-by-step answer :
According to the problem we have a box which contains a total of 10 identical items. It is said that 3 out of 10 items are found to be detective.
Now we need to find the probability that at least 2 out of 4 are found to be defective if 4 items are drawn randomly without replacement.
We know that probability of an event is defined as follows:
$\text{p(event)=}\dfrac{\text{Total no}\text{. of favourable events}}{\text{Total no}\text{. of events occured}}$
Since, it is said that we need at least 2 out of 4 are found to be defective. So, we find the probabilities for 2 out of 4 items found to be defective and 3 out of 4 to be defective cases separate and we add them later.
We first find the probability for which 2 out of 4 are found to be defective. Let us assume that ‘p’ is the probability of drawing defective items in that draw and ‘q’ be the probability for drawing non-defective items in that draw.
P(2 out of 4) = ppqq + pqpq + pqqp + qppq + qpqp + qqpp
Here $\text{p=}\dfrac{\text{no}\text{. of defective items drawn}}{\text{total no}\text{. of items}}$ and $\text{q=}\dfrac{\text{no}\text{. of non-defective items drawn}}{\text{total no}\text{. of items drawn}}$ .
$P\text{(2 out of 4)}=\left( \dfrac{3}{10}\times \dfrac{2}{9}\times \dfrac{7}{8}\times \dfrac{6}{7} \right)+\left( \dfrac{3}{10}\times \dfrac{7}{9}\times \dfrac{2}{8}\times \dfrac{6}{7} \right)+\left( \dfrac{3}{10}\times \dfrac{7}{9}\times \dfrac{6}{8}\times \dfrac{2}{7} \right)+\left( \dfrac{7}{10}\times \dfrac{3}{9}\times \dfrac{2}{8}\times \dfrac{6}{7} \right)+\left( \dfrac{7}{10}\times \dfrac{3}{9}\times \dfrac{6}{8}\times \dfrac{2}{7} \right)+\left( \dfrac{7}{10}\times \dfrac{6}{9}\times \dfrac{3}{8}\times \dfrac{2}{7} \right)$
$P\text{(2 out of 4)}=\left( \dfrac{252}{5040} \right)+\left( \dfrac{252}{5040} \right)+\left( \dfrac{252}{5040} \right)+\left( \dfrac{252}{5040} \right)+\left( \dfrac{252}{5040} \right)+\left( \dfrac{252}{5040} \right)$
$P\text{(2 out of 4)}=\left( \dfrac{252\times 6}{5040} \right)$
$P\text{(2 out of 4)}=\left( \dfrac{1512}{5040} \right)---(1)$
We first find the probability for which 3 out of 4 are found to be defective. Let us assume that ‘p’ is the probability of drawing defective items in that draw and ‘q’ be the probability for drawing non-defective items in that draw.
P(2 out of 4) = pppq + ppqp + pqpp + qppp
Here $\text{p=}\dfrac{\text{no}\text{. of defective items drawn}}{\text{total no}\text{. of items}}$ and $\text{q=}\dfrac{\text{no}\text{. of non-defective items drawn}}{\text{total no}\text{. of items drawn}}$ .
$P\text{(3 out of 4)=}\left( \dfrac{3}{10}\times \dfrac{2}{9}\times \dfrac{1}{8}\times \dfrac{7}{7} \right)+\left( \dfrac{3}{10}\times \dfrac{2}{9}\times \dfrac{7}{8}\times \dfrac{1}{7} \right)+\left( \dfrac{3}{10}\times \dfrac{7}{9}\times \dfrac{2}{8}\times \dfrac{1}{7} \right)+\left( \dfrac{7}{10}\times \dfrac{3}{9}\times \dfrac{2}{8}\times \dfrac{1}{7} \right)$
$P\text{(3 out of 4)=}\left( \dfrac{42}{5040} \right)+\left( \dfrac{42}{5040} \right)+\left( \dfrac{42}{5040} \right)+\left( \dfrac{42}{5040} \right)$
$P\text{(3 out of 4)=}\left( \dfrac{42\times 4}{5040} \right)$
$P\text{(3 out of 4)=}\left( \dfrac{168}{5040} \right)---(2)$
Now, we find the probability for which at least 2 out of 4 are found to be defective.
P(at least 2 out of 4) = P(2 out of 4) + P(3 out of 4)
$P\text{(atleast 2 out of 4)=}\left( \dfrac{168}{5040} \right)+\left( \dfrac{1512}{5040} \right)$
$P\text{(atleast 2 out of 4)=}\left( \dfrac{1680}{5040} \right)$
$P\text{(atleast 2 out of 4)=}\left( \dfrac{1}{3} \right)$
∴ The probability that at least 2 out of 4 items are found to be detective if 4 items are drawn without replacement is $\dfrac{1}{3}$ .
So, the correct answer is “Option D”.

Note : We should all 6 cases while calculating 2 out of 4 found to be defective and all 4 cases while calculating the 3 out of 4 found to be detective. Because order is important while drawing items from a bag. Even though probability is the same for all such cases but final probability differs which results in the wrong answer finally.