# Easy Probability Solved QuestionAptitude Discussion

 Q. From a bag containing 4 white and 5 black balls a man drawn 3 balls at random. What are the odds against these balls being black?
 ✖ A. 5/37 ✔ B. 37/5 ✖ C. 11/13 ✖ D. 13/37

Solution:
Option(B) is correct

Probability of all three balls being black

$=\dfrac{^5C_3}{^9C_3}$

$=\dfrac{5}{42}$

Probability that three balls are not black

$=1-\dfrac{5}{42}$

$=\dfrac{37}{42}$

Hence, odds against these ball being black

$=\left(37/42\right):\left(5/42\right)$

$=37:5$

## (2) Comment(s)

Sudhanva Dixit
()

Firstly they have asked the odds 'against', that means that (1-P(all Black)).

Coming to your question, It means that in 37/42 cases all the balls are not black. The probability of 2 black & 1 white, 1 Black & 2 white and all three white combined makes 37/42.

VeeraraghavanK
()

Probability that three balls are not black is given as 37/42. It means that the potability that all three balls are white is 37/42. How it is possible?

Since the number of white balls are less than black balls, its probability should be less. Also, calculating the probability of all balls being white = 1/21.

Can some one explain.