If 12 identical balls are to be placed in 3 different boxes, then the probability that one of the boxes contains excatly 3 balls, is

Question

If 12 identical balls are to be placed in 3 different boxes, then the probability that one of the boxes contains excatly 3 balls, is (A) (55/3)((2/3))11 (B) 55((2/3))10 (C) 220((1/3))12 (D) 22((1/3))11

Tardigrade Admin · Accepted Answer

Success (p)=(1/3) Failure (q)=(2/3) Acc. to binomial distribution, we have to find, P ( x =3)=12 C 3 ⋅((1/3))3 ⋅((2/3))9 =(55/3)((2/3))11

Q. If $12$ identical balls are to be placed in $3$ different boxes, then the probability that one of the boxes contains excatly $3$ balls, is

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

$55 (\frac{2}{3})^{10}$

$220 (\frac{1}{3})^{12}$

$22 (\frac{1}{3})^{11}$

Success $(p) = \frac{1}{3}$
Failure $(q) = \frac{2}{3}$
Acc. to binomial distribution, we have to find,
$P (x = 3) = 1 2_{C_{3}} \cdot (\frac{1}{3})^{3} \cdot (\frac{2}{3})^{9}$
$= \frac{55}{3} (\frac{2}{3})^{11}$

Q. If 12 identical balls are to be placed in 3 different boxes, then the probability that one of the boxes contains excatly 3 balls, is

355​(32​)11

55(32​)10

220(31​)12

22(31​)11