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probability lottery question ??

Expert replies
by raunakrajan » Wed Jun 30, 2010 3:30 am
how to solve this one!?

For a lottery, three numbers are drawn, each from 1 to 40. A player can win if the product of his or her three numbers is odd. Initially, the same number may be picked up to three times. Later, the rules are changed so that each number may only be picked once. Approximately how much does this reduce a person's probability of winning?

0.01
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Source: — Problem Solving |

by kvcpk » Wed Jun 30, 2010 4:09 am
raunakrajan wrote:how to solve this one!?

For a lottery, three numbers are drawn, each from 1 to 40. A player can win if the product of his or her three numbers is odd. Initially, the same number may be picked up to three times. Later, the rules are changed so that each number may only be picked once. Approximately how much does this reduce a person's probability of winning?

0.01
product of three numbers is odd only when those numbers are odd.
We have 20 odd numbers between 1 to 40.

So prob of winning with first ruleset is (20*20*20)/(40*40*40) [because numbers can be repeated]

Second time, we can choose one number only once. So,
prob is (20*19*18)/(40*39*38)

we get 1/8 - 9/78 = approx.. 1/8-9/80 = 0.01
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