A bag contains 15 tickets, number from 1 to 15. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that tickets will show even numbers.
A) 1/5
B) 3/5
C) 4/5
D) 13/20
E) 2/5
GMAT Question of the Day
Subscribe / Share
looks like the question assumes that 2 is not considered as even ( it is actually a prime number). That leaves us with choice B ( 6 even numbers between 1 and 15)
I guess the answer options submitted arewrong. 2 is always considered to be an even no. and hence answer should be 7/15
Dear Sumit,
Thanks for reminding mistake. We made appropriate correction. Extremely sorry for inconvenience.
Admin
takegmat.com
hi admin,
pls explain me the ans provided by sumit, i think “A” should b the correct one…
The probability that 1st ticket picked up is even: 7/15
The probability that 2nd ticket picked up is even (assuming 1st ticket was odd): (8/15)*(7/14)=4/15
The probability that 2nd ticket is even: (7/15)*(6/14)=3/15
Hence, the overall probability of tickets showing even is: 4/15 + 3/15 = 7/15. That is choice B.
Please correct me.
But choice B is 3/5.
I referred to Sumit’s comment and assumed that choice B is wrong and the correct value for choice B is 7/15 which turns out to be an answer.
Let us know the correct answer for this post.
well in the first case the probability of pickin up 1 even ticket from 15 tickets will be 7c1/15c1=7/15.
in the next case since the ticket is not replaced ..the total number of tickets now in the bag will be 14 and the total number of even numbers now will be 6..considering the first num picked is even..
so probability now is..6c1/14c1=6/14.
therefore..ans wud be…(7/15)*(6/14)=(1/5)
( Coountributed by Avinash)
A
I beleive the answer should be 1/5 , A
answer is
7c1* 6c1/ 15c1* 14c1 = 1/5
A) 1/5
7c1 * 6c1
———– = 1/5
15c1*14c1
A is correct.
A
A
ans A
My approach
first ticket=7/15
second=6/14 as we have to consider that the first pull was an even number
then…
(7/15)*(6/14)
Ans: 1/5
I think the question is flawed because it needs to specify whether the first picked turns out to be even or odd, or allow for two answers because if the first number picked is odd then the overall probability that the two numbers picked are even is 7/30. If the first number picked turns out to be even, the overall probability is 1/5. Now since 7/30 isn’t an answer choice, if you were taking the test you could assume 1/5 was the best answer and so choose it but technically I think the answer depends on the outcome of the first ticket chosen.
the question is unambiguous. It asks for BOTH tickets showing even numbers. Hence A.
I think the question is asking if both the tickets are even numbered. As the question says:
A bag contains 15 tickets, number from 1 to 15. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that “tickets will show even numbers”.
Please see the last part of the question. Correct me if I am wrong.
According to me the answer is 1/5. which is option A.
7/15 * 6/14 = 1/5
we have total 15 tickets and 7 r even numbered & 8 r odd numbered.
probability of finding two even numbered tickets in 15 tickets
= 7c2/15c2 = 7*6/15*14 = 1/5 ans
I think there is no flaw in the question
it asks probability of both the tickets being even
why we should consider Ist being odd? that should not come in counting
and it 7/15*6/14 as the events are in succession and without replacement
Please correct me if I am wrong
Ans A
A= Probability of A.Probability of B = P(A.B) of successive independent events.
A