GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A hat contains 18 raffle tickets, numbered 1 through 18. If
[#permalink]
24 Nov 2017, 20:43
24 Nov 2017, 20:43
Expert Reply
00:00
Question Stats:
76% (01:11) correct
23% (00:37) wrong based on 56 sessions
Hide Show timer Statistics
A hat contains 18 raffle tickets, numbered 1 through 18. If two raffle tickets are chosen at random from the hat, what is the probability that both tickets are even numbers?
A. \(\frac{2}{9}\)
B. \(\frac{4}{17}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{33}{34}\)
A. \(\frac{2}{9}\)
B. \(\frac{4}{17}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{33}{34}\)
Drill 1
Question: 8
Page: 522
Question: 8
Page: 522
Re: A hat contains 18 raffle tickets, numbered 1 through 18. If
[#permalink]
02 Dec 2017, 10:41
02 Dec 2017, 10:41
1
sandy wrote:
A hat contains 18 raffle tickets, numbered 1 through 18. If two raffle tickets are chosen at random from the hat, what is the probability that both tickets are even numbers?
A. \(\frac{2}{9}\)
B. \(\frac{4}{17}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{33}{34}\)
A. \(\frac{2}{9}\)
B. \(\frac{4}{17}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{33}{34}\)
Drill 1
Question: 8
Page: 522
Question: 8
Page: 522
Here
In the number from 1 to 18,
number of even numbers = 9
number of odd numbers = 9
Therefore probability of picking two even numbers = \(\frac{9}{18} * \frac{8}{17}\) = \(\frac{4}{17}\)
Re: A hat contains 18 raffle tickets, numbered 1 through 18. If
[#permalink]
10 Dec 2017, 13:17
10 Dec 2017, 13:17
Expert Reply
Explanation
Think of this problem as if you’re pulling out an even ticket and then another even ticket.
So, for the first ticket there are 9 possible evens out of 18 total, so the probability that the first ticket is even is \(\frac{9}{18}\). Now you have one fewer even ticket in the hat. So there are 8 evens out of 17 total tickets for the second ticket, thus, the probability is \(\frac{8}{17}\). You want an even AND aneven, so multiply: \(\frac{9}{18}\times\frac{8}{17}=\frac{4}{17}\).
The answer is choice B.
Think of this problem as if you’re pulling out an even ticket and then another even ticket.
So, for the first ticket there are 9 possible evens out of 18 total, so the probability that the first ticket is even is \(\frac{9}{18}\). Now you have one fewer even ticket in the hat. So there are 8 evens out of 17 total tickets for the second ticket, thus, the probability is \(\frac{8}{17}\). You want an even AND aneven, so multiply: \(\frac{9}{18}\times\frac{8}{17}=\frac{4}{17}\).
The answer is choice B.
