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probability

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by srivathsan » Sat Sep 08, 2012 7:26 am
there are 100 cards numbered from 1 to 100 . If three cards are drawn at random and with replacement, what is the probability that the sum of three numbers on the cards so selected will be odd?

a.1/4
b.3/8
c.1/2
d.5/8
e.3/4
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Source: — Problem Solving |

by vk_vinayak » Sat Sep 08, 2012 8:23 am
srivathsan wrote:there are 100 cards numbered from 1 to 100 . If three cards are drawn at random and with replacement, what is the probability that the sum of three numbers on the cards so selected will be odd?

a.1/4
b.3/8
c.1/2
d.5/8
e.3/4
Not sure if this method is correct.

Two possibilities will make the sum odd: All three are ODD -or- two ODD and one even.

Possibility of all three odd: (50/100) * (50/100) *(50/100) = 1/8
Possibility of two odd and one even: (50/100) * (50/100) *(50/100) = 1/8

Total: (1/8)+(1/8) = 1/4. Choise A.
- VK

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