Let six children $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ each have a fair 6 sided die which they roll simultaneously. Let E denotes the event that atleast three of them roll a prime number and F denotes the event that $B_{1}, B_{2}, B_{3}$ throw a prime number. If the conditional probability of F given E is equal to $\frac{a}{b}$ where a and b are coprime then which is/are CORRECT ?

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$b - 4 a = 0$

$| a - b | = 10$

$ab \leq 84$

$2 a + b = 29$

answer is A, C.

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Detailed Solution

P(Prime number) $= \frac{1}{2}$
$\begin{array}{l} ∴ p = \frac{1}{2}; q = \frac{1}{2}; n = 6 \\ P (E) = P (X \geq 3) \\ = \frac{^{6} C_{3} +^{6} C_{4} +^{6} C_{5} +^{6} C_{6}}{64} \\ = \frac{42}{64} = \frac{21}{32} \\ P (F) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} \\ P (F / E) = \frac{P (F \cap E)}{P (E)} = \frac{P (F)}{P (E)} \\ = \frac{1}{8} (\frac{32}{21}) = \frac{4}{21} = \frac{a}{b} \\ ∴ a = 4 and b = 21 \end{array}$