Question

A bag contains 6 red and 9 blue balls. Two successive drawing of four balls are made such that the balls are not replaced before the second draw. Find the probability that the first draw gives 4 red balls and second draw gives 4 blue balls.

• A. $\frac{3}{715}$
• B. $\frac{7}{715}$
• C. $\frac{15}{233}$
• D. None of these

Answer 1

The correct option is A $\frac{3}{715}$

Let A be the event drawing 4 red ball in first draw and B be the event of drawing 4 blue balls in the second draw.
Then $P\left(A\right)=\frac{{}^6{C}_4}{{}^{15}{C}_4}=\frac{5}{1365}=\frac{1}{91}$
$P\left(\frac{B}{A}\right)=\frac{{}^9{C}_4}{{}^{11}{C}_4}=\frac{126}{330}=\frac{21}{55}$
Hence, the required probability = $P(A\cap B)$
$=P\left(A\right)P\left(\frac{B}{A}\right)$
$=\frac{1}{91}\times \frac{21}{55}=\frac{3}{715}$