JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Probability

A fair die is tossed until six is obtained on it. Let

X

be the number of required tosses, then the conditional probability

P (X \geq 5 ∣ X > 2)

is:

Answer: (B)

Solution

The event

X > k

means the first

k

tosses all failed to show a six, so

P (X > k) = (\frac{5}{6})^{k}

The event

X \geq 5

is the same as

X > 4

, and it is contained in

X > 2

. Hence

P (X \geq 5 ∣ X > 2) = \frac{P ( X > 4 )}{P ( X > 2 )} = \frac{( 5/6 ) ^{4}}{( 5/6 ) ^{2}} = (\frac{5}{6})^{2} = \frac{25}{36}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option