If 12 identical balls are to be placed in 3 d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The total number of ways to place $12$ identical balls into $3$ distinct boxes is given by the stars and bars formula: $\binom{12+3-1}{3-1} = \bin

Vedclass · Accepted Answer

$\frac{55}{3}{\left( {\frac{2}{3}} \right)^{11}}$

Vedclass · Answer

(B) The total number of ways to place $12$ identical balls into $3$ distinct boxes is given by the stars and bars formula: $\binom{12+3-1}{3-1} = \binom{14}{2} = 91$. However,in probability problems involving placing balls into boxes,we typically treat the balls as being placed independently into boxes with probability $p = \frac{1}{3}$ for each box.Let $X$ be the number of balls in a specific box. $X$ follows a binomial distribution $B(n, p)$ where $n = 12$ and $p = \frac{1}{3}$.The probability that a specific box contains exactly $3$ balls is $P(X = 3) = \binom{12}{3} \left(\frac{1}{3}ight)^3 \left(\frac{2}{3}ight)^9$.Since there are $3$ boxes,the probability that at least one box contains exactly $3$ balls is calculated using the binomial distribution logic for the selection of boxes.$P = \binom{12}{3} \left(\frac{1}{3}ight)^3 \left(\frac{2}{3}ight)^9 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} 	imes \frac{1}{27} 	imes \left(\frac{2}{3}ight)^9 = 220 	imes \frac{1}{27} 	imes \left(\frac{2}{3}ight)^9 = \frac{220}{27} 	imes \left(\frac{2}{3}ight)^9$.Simplifying,$\frac{220}{27} = \frac{55 	imes 4}{27} = \frac{55}{3} 	imes \frac{4}{9} = \frac{55}{3} 	imes \left(\frac{2}{3}ight)^2$.Thus,$P = \frac{55}{3} 	imes \left(\frac{2}{3}ight)^2 	imes \left(\frac{2}{3}ight)^9 = \frac{55}{3} \left(\frac{2}{3}ight)^{11}$.

If $12$ identical balls are to be placed in $3$ distinct boxes,then the probability that one of the boxes contains exactly $3$ balls is:

Similar Questions

If the mean and variance of a binomial variate $X$ are $8$ and $4$ respectively,then $P(X < 3)$ equals to

The mean and the variance of a binomial distribution are $4$ and $2$ respectively. Then the probability of $2$ successes is

Suppose $X$ has a binomial distribution $B(6, 1/2)$. Show that $X=3$ is the most likely outcome.
(Hint: $P(X=3)$ is the maximum among all $P(x_i)$,where $x_i = 0, 1, 2, 3, 4, 5, 6$)

If the sum of mean and variance of a Binomial Distribution is $\frac{15}{2}$ for $10$ trials,then the variance is (in $.5$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module