Each of the nine balls can be placed with equal probability in any of the three initially empty boxes. Let P1 be the probability that there will be four balls in the first box, three in the second box and two in the third box, and P 2 be the probability that there will be three balls in each box then the ratio ( P 2 / P 1) equals

Question

Each of the nine balls can be placed with equal probability in any of the three initially empty boxes. Let P1 be the probability that there will be four balls in the first box, three in the second box and two in the third box, and P 2 be the probability that there will be three balls in each box then the ratio ( P 2 / P 1) equals (A)  (4/3) (B) (2/9) (C) 8 (D) 2

Tardigrade Admin · Accepted Answer

n ( S )=39 P2=(9 ! ⋅ 3 !/(3 !)3 ⋅ 3 !) ⋅ (1/39) (all boxes are different) P1=(9 !/4 ! ⋅ 3 ! ⋅ 2 !) ⋅ (1/39) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/09def20b31d6703d9cbc4708cd05131b-.png" /> (P2/P1)=(9 !/(3 !)3) ⋅ (4 ! ⋅ 3 ! ⋅ 2 !/9 !)=(4/3)

$\frac{4}{3}$

$\frac{2}{9}$

8

2

$n (S) = 3^{9}$
$P_{2} = \frac{9 ! \cdot 3 !}{( 3 ! ) ^{3} \cdot 3 !} \cdot \frac{1}{3 ^{9}}$
(all boxes are different)
$P_{1} = \frac{9 !}{4 ! \cdot 3 ! \cdot 2 !} \cdot \frac{1}{3 ^{9}}$

$\frac{P _{2}}{P _{1}} = \frac{9 !}{( 3 ! ) ^{3}} \cdot \frac{4 ! \cdot 3 ! \cdot 2 !}{9 !} = \frac{4}{3}$

34​

92​

8

2

n(S)=39 P2​=(3!)3⋅3!9!⋅3!​⋅391​ (all boxes are different) P1​=4!⋅3!⋅2!9!​⋅391​ P1​P2​​=(3!)39!​⋅9!4!⋅3!⋅2!​=34​

$\frac{4}{3}$

$\frac{2}{9}$

$n (S) = 3^{9}$
$P_{2} = \frac{9 ! \cdot 3 !}{( 3 ! ) ^{3} \cdot 3 !} \cdot \frac{1}{3 ^{9}}$
(all boxes are different)
$P_{1} = \frac{9 !}{4 ! \cdot 3 ! \cdot 2 !} \cdot \frac{1}{3 ^{9}}$

$\frac{P _{2}}{P _{1}} = \frac{9 !}{( 3 ! ) ^{3}} \cdot \frac{4 ! \cdot 3 ! \cdot 2 !}{9 !} = \frac{4}{3}$