A basket contains $20$ apples and $10$ oranges out of which $5$ apples and $3$ oranges are defective. If a person takes out $2$ at random, then the probability that either both are apples or both are good is $\frac{316}{435}$ .

True

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False

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Solution

The correct option is A True
Out of 30 items, two can be selected in $^{30} C_{2}$ ways.
So total number of elementary event $^{30} C_{2}$
Consider the following events:
$A =$ Getting two apples;
$B =$ Getting two good items
Required probability
$= P (A \cup B) = P (A) + P (B) - P (A \cap B) . . . . (i)$
There are 20 apples, out of which 2 can be drawn in $^{30} C_{2}$ ways.
$∴ P (A) = \frac{^{20} C_{2}}{^{30} C_{2}}$
There are 8 defective pieces and the remaining 22 are good. Out of 22 good pieces, two can be selected in $^{22} C_{2}$ ways
$∴ P (B) = \frac{^{22} C_{2}}{^{30} C_{2}}$
Since there are 15 pieces which are good apples out of which 2 can be selected in $^{15} C_{2}$ ways. Therefore, $P (A \cap B) =$ Probability of getting 2 pieces which are good apples
$= \frac{^{15} C_{2}}{^{30} C_{2}}$
Now,
Required probability
$= P (A) + P (B) - (A \cap B)$ [from (i)]
$= \frac{^{20} C_{2}}{^{30} C_{2}} + \frac{^{22} C_{2}}{^{30} C_{2}} - \frac{^{15} C_{2}}{^{30} C_{2}} = \frac{316}{435}$

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A basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges are defective. If a person takes out 2 at random, then the probability that either both are apples or both are good is 316435.

A basket contains $20$ apples and $10$ oranges out of which $5$ apples and $3$ oranges are defective. If a person takes out $2$ at random, then the probability that either both are apples or both are good is $\frac{316}{435}$ .