Q). In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?
A) 8/25
B) 12/35
C) 13/35
D) 9/25
E) 17/25
OA is B
I did the question in this way and was not able to get to the right answer:-
I found the probability of getting a nickel in the first 2 shots and subtract it from 1
i.e.
1st pick: 6/15
2nd pick: 5/14
therefore, 6*5/(15*14)
-> 1/7
therefore probability of not getting a nickel is 1- 1/7=> 6/7
But I see that I'm doing something wrong here.
So,I had to move back to the conventional method in order to get the answer
i.e.
P(D)*P(D) + P(P)*P(P) +2 P(D)P(P)
Can someone explain me, where I went wrong with my earlier approach.
~Thanks,
Akshat
A) 8/25
B) 12/35
C) 13/35
D) 9/25
E) 17/25
OA is B
I did the question in this way and was not able to get to the right answer:-
I found the probability of getting a nickel in the first 2 shots and subtract it from 1
i.e.
1st pick: 6/15
2nd pick: 5/14
therefore, 6*5/(15*14)
-> 1/7
therefore probability of not getting a nickel is 1- 1/7=> 6/7
But I see that I'm doing something wrong here.
So,I had to move back to the conventional method in order to get the answer
i.e.
P(D)*P(D) + P(P)*P(P) +2 P(D)P(P)
Can someone explain me, where I went wrong with my earlier approach.
~Thanks,
Akshat
Last edited by akshatgupta87 on Sun Apr 01, 2012 10:42 am, edited 1 time in total.
