JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Probability XII

Consider three sets

E_{1} = {1, 2, 3}, F_{1} = {1, 3, 4}

and

G_{1} = {2, 3, 4, 5}

. Two elements are chosen at random, without replacement, from the set

E_{1}

, and let

S_{1}

denote the set of these chosen elements.
Let

E_{2} = E_{1} - S_{1}

and

F_{2} = F_{1} \cup S_{1}

. Now two elements are chosen at random, without replacement, from the set

F_{2}

and let

S_{2}

denote the set of these chosen elements.
Let

G_{2} = G_{1} \cup S_{2}

. Finally, two elements are chosen at random, without replacement, from the set

G_{2}

and let

S_{3}

denote the set of these chosen elements.
Let

E_{3} = E_{2} \cup S_{3}

. Given that

E_{1} = E_{3}

, let p be the conditional probability of the event

S_{1} = {1, 2}

. Then the value of

p

Answer: (A)

Solution

We need

P = \frac{P ( S _{1} = { 1 , 2 } \cap ( E _{1} = E _{3} ))}{P ( E _{1} = E _{3} )} = \frac{P ( B _{1, 2} )}{P ( B )}

. For

S_{1} = {1, 2}

P (B_{1, 2}) = \frac{1}{3} \times \frac{1 \times ^{3} C _{1}}{^{4} C _{2}} \times \frac{1}{^{5} C _{2}}

(1 is definitely chosen from

F_{2}

, and both 1,2 must be chosen from

G_{2}

). Similarly compute

P (B_{1, 3})

and

P (B_{2, 3})

for the other possible pairs. The total probability

P (B) = P (B_{1, 2}) + P (B_{1, 3}) + P (B_{2, 3})

. Computing the ratio gives

\frac{P ( B _{1, 2} )}{P ( B )} = \frac{1}{5}

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Single correct option