Question

A bag contains 3 red, 6 white and 7 blue balls. Two balls are drawn one by one. What is the probability that first ball is white and second ball is blue where first drawn ball is not replaced in the bag?

• A. $\frac{7}{40}$
• B. $\frac{9}{40}$
  
• C. $\frac{11}{40}$
  
• D. $\frac{3}{40}$
  

Answer 1

The correct option is A $\frac{7}{40}$

Let A = Event of drawing a white ball in first draw
and B = Event of drawing a blue ball in second draw.
Here A and B are dependent events.
$P\left(A\right)=\frac{6}{16},P\left(\frac{B}{A}\right)=\frac{7}{15}$
$P(A\cap B)=P\left(A\right).P\left(\frac{B}{A}\right)=\frac{6}{16}\times \frac{7}{15}=\frac{7}{40}$.