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01 Oct 2017, 12:48
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B
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A bowl contains 18 raffle tickets, numbered 1 through 18. If two raffle tickets are chosen at random from the bowl, 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. \(\frac{33}{34}\)
01 Oct 2017, 20:17
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Gnpth
Hi...
There are 9 even numbers out of 18 consecutive integers..
So Q boils down to selecting 2 out of 9... 9P2=\(9!/7!=8*9\)
And total ways are 18*17..
So prob= \(\frac{8*9}{17*18}=4/17\)
B
Also if you work directly on probability..
Prob of picking an even number in first go= 9/18=1/2
Prob of picking an even number then on second go also =1/2 * 8/17=4/17
B
02 Oct 2017, 09:07
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Gnpth
Even numbers are 9 then probability 1st even raffle is 9/18 and for second is 8/17
=9/18*8/17=4/17.. Ans B
