probability query

Question

hi,please xplain
1) A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16
ans A

2)My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
ans B

tishan · Answer

3)How many 5-digit positive integers exist where no two consecutive digits are the same? A.) 9*9*8*7*6 B.) 9*9*8*8*8 C.) 9^5 D.) 9*8^4 E.) 10*9^4 Soln: C is correct.

lachlanc · Answer

Question 1 Answer - probability that both not blue = 6/8 * 5/7 = 30/56 = 15/28 ie - there are 6 of 8 which are not blue on the first draw and on the next draw, there are 5 of 7 which are not blue.

tishan · Answer

thanks lachlanc...
but one thing we why cant we find the probability of picking up both the blues and then subtract it from 1....
1-(2/8 *1/7)
but the ans in wrong in this case
please explain where i am going wrong.

tishan

mostovari · Answer

Did you get the answer for your other two questions? Please let us know

sam77 · Answer

tishan wrote:thanks lachlanc...
but one thing we why cant we find the probability of picking up both the blues and then subtract it from 1....
1-(2/8 *1/7)
but the ans in wrong in this case
please explain where i am going wrong.

tishan

I made the same mistake. But apparently, we have misread the question. It asks for probablity that both are not blue, that means EITHER one of them is blue and another non-blue OR both are non-blue. But, what we calculated was the probablity of both being non-blue.
Also, what are the choices for your 2nd question. I am getting 1/60 for AJEET. Total Cases = Factorial(5)/2. Favourable cases = 1.

cramya · Answer

Tishan/Sam,

I dont think they are complementary event

If u were to do that it would be

1 - (p(bothblue) + p(1st one not blue 2nd one blue) + p(1stone blue 2ndone not blue)

1- (2/8*1/7 + 6/8*2/7 + 2/8*6/7)
1- (2/56+12/56+12/56)
1- (26/56)
=30/56
= 15/28

Therefore its easier to do
p(notblue1st pick notblue 2ndpick) = 6/8*5/7 = 30/56 = 15/28

earth@work · Answer

sam77 wrote:tishan wrote:thanks lachlanc... but one thing we why cant we find the probability of picking up both the blues and then subtract it from 1.... 1-(2/8 *1/7) but the ans in wrong in this case please explain where i am going wrong. tishan I made the same mistake. But apparently, we have misread the question. It asks for probablity that both are not blue, that means EITHER one of them is blue and another non-blue OR both are non-blue. But, what we calculated was the probablity of both being non-blue. Also, what are the choices for your 2nd question. I am getting 1/60 for AJEET. Total Cases = Factorial(5)/2. Favourable cases = 1. Well,count me as well for the mistake in 1st question. For the 2nd question i m also getting 1/60, my method was same as Sam. what is the OA 3rd question: 5 digit positive integer 9(Excluding zero)*9(incl zero)*9(incl zero)*9(incl zero)*9(incl zero)=9^5

cramya · Answer

I think the key to question 3 is to realize the numbers before the number u just picked comes back in to the pool for selection of the one u r about to pick making it 9^5

Earth@work already explained the inclusion/exclusion of 0's

cramya · Answer

I think the key to question 3 is to realize the numbers before the number u just picked comes back in to the pool for selection FOR the one u r about to pick making it 9^5

Earth@work already explained the inclusion/exclusion of 0's

jimmiejaz · Answer

cramya wrote:Tishan/Sam,

I dont think they are complementary event

If u were to do that it would be

1 - (p(bothblue) + p(1st one not blue 2nd one blue) + p(1stone blue 2ndone not blue)

1- (2/8*1/7 + 6/8*2/7 + 2/8*6/7)
1- (2/56+12/56+12/56)
1- (26/56)
=30/56
= 15/28

Therefore its easier to do
p(notblue1st pick notblue 2ndpick) = 6/8*5/7 = 30/56 = 15/28

Hi Cramya,

Please explain this again.The question asks to find the probability when both are not blue. Doesnt That mean one can be blue, or both can be any color other than blue. what i think is that the ans should be 13/28 or 26/56 as u hv found.
Am really confused in this..

cramya · Answer

Hi Cramya,

Please explain this again.The question asks to find the probability when both are not blue. Doesnt That mean one can be blue, or both can be any color other than blue. what i think is that the ans should be 13/28 or 26/56 as u hv found.
Am really confused in this..

The easiest approach would be

P(not blue in 1st draw) * p (not blue in the 2nd draw)

= 6/8 *5/7
= 30/ 56
= 15/28

The other way I did it to show Tishan the 1 - (complementary method) way.

logitech · Answer

okay

8C2 = 28 different ways

We have two blue cards, B1 and B2 and 6 other cards

so we can have B1(something) , B2(something), or B1B2

B1something = 1 x 6c1 = 6
B1something = 1 x 6c1 = 6
B1B2 =1c1 =1

So 13 of the ways have BLUE in it so

(28-13)/28

15/28

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

jimmiejaz · Answer

thanks a lot logitech. I got it. well i think i over analyzed the problem. Thanks for pointing out. Btw, when are you giving the GMAT?

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