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A hat contains 18 raffle tickets, numbered 1 through 18. If
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24 Nov 2017, 20:43
24 Nov 2017, 20:43
Expert Reply
00:00
Question Stats:
75% (01:07) correct
24% (00:37) wrong based on 54 sessions
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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
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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
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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.
