(B) Total number of balls $= 4 + 5 + 6 = 15$.We need to draw $3$ balls of different colours, which means one red, one blue, and one yellow ball.The nu

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(B) Total number of balls $= 4 + 5 + 6 = 15$.We need to draw $3$ balls of different colours, which means one red, one blue, and one yellow ball.The number of ways to choose $1$ red, $1$ blue, and $1$ yellow ball is $\binom{4}{1} \times \binom{5}{1} \times \binom{6}{1} = 4 \times 5 \times 6 = 120$.The total number of ways to draw $3$ balls from $15$ is $\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.The required probability is $\frac{120}{455} = \frac{24}{91}$.

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

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$3$ balls are drawn one after the other without replacement from an urn containing $4$ red, $5$ blue and $6$ yellow balls. The probability of getting three different coloured balls is

TS EAMCET 2022 All TS EAMCET

A
$\frac{12}{91}$
B
$\frac{24}{91}$
C
$\frac{8}{225}$
D
$\frac{8}{75}$

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Advanced Use of permutations and combinations in probability

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