JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Probability

In a tournament, a team plays 10 matches with probabilities of winning and losing each match as

\frac{1}{3}

and

\frac{2}{3}

respectively. Let

x

be the number of matches that the team wins, and

y

be the number of matches that they lose. If the probability

P (∣ x - y ∣ \leq 2)

p

, then

3^{9} p

equals ____.

Answer: 8288

Solution

P (win) = \frac{1}{3}, P (loss) = \frac{2}{3}

, 10 matches,

x + y = 10

. Need

P (∣ x - y ∣ \leq 2)

, i.e.

x - y \in {- 2, - 1, 0, 1, 2}

; since

x + y = 10

is even,

x - y

is even, so only

x - y \in {- 2, 0, 2}

are possible, giving

(x, y) \in {(4, 6), (5, 5), (6, 4)}

p = (5 10) (\frac{1}{3})^{5} (\frac{2}{3})^{5} + (6 10) (\frac{1}{3})^{6} (\frac{2}{3})^{4} + (4 10) (\frac{1}{3})^{4} (\frac{2}{3})^{6}

Each term has

\frac{1}{3}^{10}

in its denominator (since the two exponents in every term add to 10), so

3^{10} p = (5 10) 2^{5} + (6 10) 2^{4} + (4 10) 2^{6} = 252 (32) + 210 (16) + 210 (64) = 8064 + 3360 + 13440 = 24864

3^{9} p = \frac{24864}{3} = 8288

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Integer answer