Let 9 distinct balls be distributed among 4 boxes, B1, B2, B3 and B4. If the probability than B3 contains exactly 3 balls is k( 3 \over 4 )^9 then k lies in the set :

Probability PYQ — JEE Main 2021 · 25 Jul · Shift 1 · Q35 with Solution

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Correct answer: A

Model the distribution

Since 9 distinct balls are distributed among 4 distinct boxes, and each ball can go to any one of the 4 boxes independently, the total number of distributions is

$4^9.$

We want the probability that box $B_3$ contains exactly 3 balls.

Count favorable distributions

To make $B_3$ contain exactly 3 balls:

Choose which 3 of the 9 distinct balls go to $B_3$ : $\binom{9}{3}.$
The remaining 6 balls must go to one of the other 3 boxes $B_1,B_2,B_4$ : $3^6$ ways.

So favorable distributions are

$\binom{9}{3}3^6.$

Hence,

$P(B_3\text{ has exactly 3 balls})=\frac{\binom{9}{3}3^6}{4^9}.$

Rewrite in the given form

We are given that this probability is

$k\left(\frac{3}{4}\right)^9.$

Now,

= \binom{9}{3}\cdot \frac{3^6}{4^9}.$$ Also, $$\left(\frac{3}{4}\right)^9 = \frac{3^9}{4^9}.$$ So, $$k\cdot \frac{3^9}{4^9} = \binom{9}{3}\cdot \frac{3^6}{4^9}.$$ Multiplying by $\frac{4^9}{3^9}$, $$k=\binom{9}{3}\cdot \frac{1}{3^3}.$$ Now, $$\binom{9}{3} = 84, \qquad 3^3=27.$$ Thus, $$k=\frac{84}{27}=\frac{28}{9}\approx 3.111\ldots$$ --- 4. **Check the options** We test $k=\frac{28}{9}$ in each set. - **A:** $|x-3|<1 \Rightarrow 2<x<4$ Since $\frac{28}{9}\approx 3.11 \in (2,4)$, this is true. - **B:** $|x-2|\le 1 \Rightarrow 1\le x\le 3$ $\frac{28}{9}>3$, so false. - **C:** $|x-1|<1 \Rightarrow 0<x<2$ False. - **D:** $|x-5|\le 1 \Rightarrow 4\le x\le 6$ False. Therefore, the correct option is **A**. --- 5. **Comparison with stored answer** Stored correct answer: **A** This matches our derived answer.

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View written solutionFree

Correct answer: A

Model the distribution

Since 9 distinct balls are distributed among 4 distinct boxes, and each ball can go to any one of the 4 boxes independently, the total number of distributions is

$4^9.$

We want the probability that box $B_3$ contains exactly 3 balls.

Count favorable distributions

To make $B_3$ contain exactly 3 balls:

Choose which 3 of the 9 distinct balls go to $B_3$ : $\binom{9}{3}.$
The remaining 6 balls must go to one of the other 3 boxes $B_1,B_2,B_4$ : $3^6$ ways.

So favorable distributions are

$\binom{9}{3}3^6.$

Hence,

$P(B_3\text{ has exactly 3 balls})=\frac{\binom{9}{3}3^6}{4^9}.$

Rewrite in the given form

We are given that this probability is

$k\left(\frac{3}{4}\right)^9.$

Now,

= \binom{9}{3}\cdot \frac{3^6}{4^9}.$$ Also, $$\left(\frac{3}{4}\right)^9 = \frac{3^9}{4^9}.$$ So, $$k\cdot \frac{3^9}{4^9} = \binom{9}{3}\cdot \frac{3^6}{4^9}.$$ Multiplying by $\frac{4^9}{3^9}$, $$k=\binom{9}{3}\cdot \frac{1}{3^3}.$$ Now, $$\binom{9}{3} = 84, \qquad 3^3=27.$$ Thus, $$k=\frac{84}{27}=\frac{28}{9}\approx 3.111\ldots$$ --- 4. **Check the options** We test $k=\frac{28}{9}$ in each set. - **A:** $|x-3|<1 \Rightarrow 2<x<4$ Since $\frac{28}{9}\approx 3.11 \in (2,4)$, this is true. - **B:** $|x-2|\le 1 \Rightarrow 1\le x\le 3$ $\frac{28}{9}>3$, so false. - **C:** $|x-1|<1 \Rightarrow 0<x<2$ False. - **D:** $|x-5|\le 1 \Rightarrow 4\le x\le 6$ False. Therefore, the correct option is **A**. --- 5. **Comparison with stored answer** Stored correct answer: **A** This matches our derived answer.

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