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probabilty problem

Problem Solving — algebra and arithmetic
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by sukh » Wed Oct 26, 2011 12:57 am
A basket contains cards bearing numbers 10, 11,... , 20. If a card is drawn from the basket and is not returned to it, and then the second card is drawn from the basket, what is the probability that the first card drawn will have a prime number on it while the second will not? (A)28/110 (B)4/15 (C)5/16 (D)27/80 (E)32/55
plz give explaination etc
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by neelgandham » Wed Oct 26, 2011 2:40 am
10,11,12,13,14,15,16,17,18,19,20 (total #'s = 11)
probability that the first card drawn will have a prime number = 4/11 ( Prime numbers in bold !), Let us say he choose 11

10,12,13,14,15,16,17,18,19,20 (total #'s = 11)
probability that the first card drawn doesn't have a prime number = 7/10

probability that the first card drawn will have a prime number on it while the second will not = (4/11)*(7/10) = [spoiler](A)28/110 [/spoiler]

Please correct me if I am wrong! Probability, and P&C are a serious threat to my score !
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by VivianKerr » Thu Oct 27, 2011 7:05 pm
There are four primes between 10 and 20: 11, 13, 17, and 19.

That's 4 out of 11 cards.

After the first is drawn, that leaves 10 cards, only 3 of which are prime.

Prob of 1st being prime: 3/11.

Prob of 2nd being non-prime: 7/10.

Probability of these events occurring together 4/11 x 7/10 = 28/110.

I agree with Neel. :)
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