If 3 apples (like) and 3 bananas (like) are distributed to three children at random then which is /are correct

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The chance that each child gets at least one fruit is $\frac{11}{20}$

The chance that all fruits go to one child is $\frac{3}{100}$

The chance that all apples go to one child is $\frac{3}{10}$

The chance that each child gets one apple and one banana is ‘P’ then sum of digits of $[\frac{1}{p}]$ is 6.

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{l} n (s) = (3 + (3 - 1) C_{3 - 1}) \times (3 + (3 - 1) C_{3 - 1}) \\ = 10 \times 10 = 100 \\ A) n (E) = n (\bar{A} \cap \bar{B} \cap \bar{C}) = n (s) - n (A \cup B \cup C) \\ = 100 - [16 + 16 + 16 - 1 - 1 + 0] = 55 \end{array}$
$B) P (E) = \frac{3}{100} = P \Rightarrow [\frac{1}{P}] = 33 sum of digits = 6$
$C) P (E) = \frac{3}{100}$
$D) P (E) = \frac{3 \times 10}{100} = \frac{3}{10}$