probability problem

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probability problem

by mrinal2100 » Tue Aug 26, 2008 3:46 am
A) Two cards are drawn at random from a wel suffled pack of cards.given than both the cards are red,what is the probability that bothe cards are:-

1)having same no
<1> 9/325 <2> 64/325 <3> 144/325 <4> 28/325


2) diff no's on them and diff suits?

<1> 72/325 <2> 81/325<3> 144/325 <4> 28/325


B)four digit no's are formed using the digits 0 to 4 without repetion.the probabillity that a no so formed is divisible by 2 is

<1> 5/8 <2> 3/8 <3> 7/9 <4>2/9


ans: for A(1)-9/325
ans-for A(2)-72/325
ans:-for B-5/8
Source: — Problem Solving |

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Re: probability problem

by parallel_chase » Tue Aug 26, 2008 4:37 am
mrinal2100 wrote:A) Two cards are drawn at random from a wel suffled pack of cards.given than both the cards are red,what is the probability that bothe cards are:-

1)having same no
<1> 9/325 <2> 64/325 <3> 144/325 <4> 28/325
2 suits - 26 red cards - 18 cards with same number

The question doesnt metion about that A, K, J, Q cannot be selected meaning this is not a GMAT question.

Probability of choosing first card = 18/26
Probability of choosing second card with same number = 1/25

total = 18/26 * 1/25 = 9/325

2) diff no's on them and diff suits?

<1> 72/325 <2> 81/325<3> 144/325 <4> 28/325
I think the answer choices are wrong.

probability of first card = 18/26
probability of different second card + different suit = 8/13

total=72/169
B)four digit no's are formed using the digits 0 to 4 without repetion.the probabillity that a no so formed is divisible by 2 is

<1> 5/8 <2> 3/8 <3> 7/9 <4>2/9
0,1,2,3,4

total number of digits that can be formed = 4*4*3*2=96
number of digits ending with 0 =4*3*2=24
number of digits ending with 2,4=3*3*2*2=36
(24+36)/96 = 5/8