In a tournament, a team plays 10 matches with probabilities of winning and losing each match as \frac13 and \frac23 respectively. Let x be the number of matches that the team wins, and y be the number of matches that team loses. If the probability P(|x-y| \leq 2) is p , then 3^9 p equals \underline\hspace2cm .

Probability PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q55 with Solution

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Correct answer: 8288

Since the team plays $10$ matches:

So, $x+y=10 \,\Rightarrow\, y=10-x.$

We need $P(|x-y|\le 2).$

Substitute $y=10-x$ : $|x-y|=|x-(10-x)|=|2x-10|.$

Thus, $|2x-10|\le 2.$

Solve: $-2\le 2x-10\le 2$ $8\le 2x\le 12$ $4\le x\le 6.$

Since $x$ is an integer, possible values are: $x=4,5,6.$

Each match is independent with:

Hence, $x\sim \text{Binomial}\left(10,\frac13\right).$

Therefore, $P(x=r)=\binom{10}{r}\left(\frac13\right)^r\left(\frac23\right)^{10-r}.$

So, $p=P(x=4)+P(x=5)+P(x=6).$

$P(x=4)=\binom{10}{4}\left(\frac13\right)^4\left(\frac23\right)^6$

=\frac{210\cdot 64}{3^{10}}.$$ $$P(x=5)=\binom{10}{5}\left(\frac13\right)^5\left(\frac23\right)^5$$ $$=252\cdot \frac{1}{3^5}\cdot \frac{2^5}{3^5} =\frac{252\cdot 32}{3^{10}}.$$ $$P(x=6)=\binom{10}{6}\left(\frac13\right)^6\left(\frac23\right)^4$$ $$=210\cdot \frac{1}{3^6}\cdot \frac{2^4}{3^4} =\frac{210\cdot 16}{3^{10}}.$$ Thus, $$p=\frac{210\cdot 64+252\cdot 32+210\cdot 16}{3^{10}}.$$ Now calculate the numerator: $$210\cdot 64=13440,$$ $$252\cdot 32=8064,$$ $$210\cdot 16=3360.$$ So, $$13440+8064+3360=24864.$$ Hence, $$p=\frac{24864}{3^{10}}.$$ 5. **Find $3^9p$** $$3^9p=3^9\cdot \frac{24864}{3^{10}}=\frac{24864}{3}=8288.$$ 6. **Final answer** $$\boxed{8288}$$ 7. **Comparison with stored answer** Stored correct answer: $8288$ This matches the derived answer.