A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{5}\]
Correct Answer: C
Solution :
Total number of pens in first bag = 4 + 2 = 6 and total number of pens in second bag = 3 + 5 = 8. The probability of selecting a white pen from first bag = \[\frac{4}{6}=\frac{2}{3}\] and probability of selecting a white pen from second bag = \[\frac{3}{8}\]. \[\therefore \] Required probability that both the pens are white =\[\frac{2}{3}\times \frac{3}{8}=\frac{1}{4}\].
You need to login to perform this action.
You will be redirected in
3 sec
