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Q.
A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, the probability of getting exactly one red ball is
Probability - Part 2
Solution:
Given that, red balls $=5$ and blue balls $=3$
Total number of balls $=5+3=8$
$P(\text { red }) =\frac{5}{8} $
$P(\text { blue }) =\frac{3}{8}$
Required probability is given by
$ P(\text { red })=P(\text { red }) P(\text { blue }) P(\text { blue })+P(\text { blue })
P(\text { red }) P(\text { blue })+P(\text { blue }) P(\text { blue }) P(\text { red })$
$=\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6}$
$=\frac{30}{336}+\frac{30}{336}+\frac{30}{336}$
$=\frac{90}{336}=\frac{15}{56}$