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Question

a bag contain $5$ blue and $4$ black balls. Three balls are drawn at random. What is the probability that $2$ are blue and $1$ black?

2016-10-30 14:34:31

pk
1 Answer

**Solution 1**

P(first ball blue, second ball blue, third ball black)

$=\dfrac{5}{9}× \dfrac{4}{8}×\dfrac{4}{7}=\dfrac{80}{504}$

P(first ball blue, second ball black, third ball blue)

$=\dfrac{5}{9}× \dfrac{4}{8}×\dfrac{4}{7}=\dfrac{80}{504}$

P(first ball black, second ball blue, third ball blue)

$=\dfrac{4}{9}× \dfrac{5}{8}×\dfrac{4}{7}=\dfrac{80}{504}$

Required probability

$=\dfrac{80}{504}+\dfrac{80}{504}+\dfrac{80}{504}=\dfrac{10}{21}$

**Solution 2**

Total number of ways in which $3$ balls can be selected

$=9\text{C}3=84$

Total number of ways in which $2$ blue balls and $1$ black ball can be selected

$=5\text{C}2×4\text{C}1=40$

Required probability $=\dfrac{40}{84}=\dfrac{10}{21}$