Probability
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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

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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.