(D) Total number of balls $= 4 + 5 + 6 = 15$. Number of ways to select $3$ balls out of $15$ is $^{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \

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(D) Total number of balls $= 4 + 5 + 6 = 15$. Number of ways to select $3$ balls out of $15$ is $^{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$. Number of ways to select $1$ white,$1$ red,and $1$ green ball is $^4C_1 \times ^5C_1 \times ^6C_1 = 4 \times 5 \times 6 = 120$. Required probability $= \frac{120}{455} = \frac{24}{91}$.

Class 11 Mathematics Probability Advanced Use of permutations and combinations in probability

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$A$ bag contains $4$ white,$5$ red and $6$ green balls. Three balls are picked up randomly. The probability that a white,a red and a green ball is drawn is

A
$\frac{15}{91}$
B
$\frac{30}{91}$
C
$\frac{20}{91}$
D
$\frac{24}{91}$

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Probability

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Advanced Use of permutations and combinations in probability

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MediumAP EAMCET 2025

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DifficultTS EAMCET 2024

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$A$ bag contains $4$ white,$5$ red and $6$ green balls. Three balls are picked up randomly. The probability that a white,a red and a green ball is drawn is

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