A bag contains 5 red and 3 blue balls are drawn at random without replacement, then the probability of getting exactly one red ball is.
A$\frac{15}{29}$
B$\frac{15}{56}$
C$\frac{45}{196}$
D$\frac{135}{392}$
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B$\frac{15}{56}$
Total balls = 5 red + 3 blue = 8
Let R be the event of getting red ball
B be the event of getting a blue ball.
Required probability = P(BBR) + R(BRB) + P(RBB)
$=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}+\frac{3}{8}\times\frac{5}{7}\times\frac{2}{6}+\frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}$
$=\frac{15}{56}$
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