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Probability problem

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by nextTarget » Fri Apr 27, 2012 7:18 am
Please help to solve the below doubt. Detailed explanation would be appreciated.

set A contains positive integers from 101 to 550, inclusive. if an integer is selected at random, what is the probability that a number with the first digit of 1,2 or 3 and last digit of 4,5 or 6 will be selected.
A. 1/3
B. 1/5
C. 1/6
D. 1/9
E. 1/15
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by Stuart@KaplanGMAT » Fri Apr 27, 2012 7:37 am
nextTarget wrote:Please help to solve the below doubt. Detailed explanation would be appreciated.

set A contains positive integers from 101 to 550, inclusive. if an integer is selected at random, what is the probability that a number with the first digit of 1,2 or 3 and last digit of 4,5 or 6 will be selected.
A. 1/3
B. 1/5
C. 1/6
D. 1/9
E. 1/15
Hi!

As always, when a common formula applies you should jot it down on your scratch paper. So:

probability = # of desired outcomes/total # of possibilities.

Total # of possibilities is simply the number of numbers, which is 450 (to count the number of consecutive integers in a set, take:

last number - first number + 1).


# of desired outcomes is the number of numbers with the properties we want, so let's start counting!

The desired blocks only include 101-199, 200-299 and 300-399. In each set of 100 there are 10 blocks of 10 (we can temporarily ignore the fact that 100 is missing since it doesn't end in 4, 5 or 6). In each block of 10, 3 numbers end in 4, 5 or 6.

So, 30 blocks of 10 * 3 = 90 numbers that match our criteria.

90 desired numbers/450 total possible numbers = 1/5 chance of choosing what we want... choose B!
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by nextTarget » Fri Apr 27, 2012 7:52 am
Great explanation Stuart.
Now its look so simple problem.. Thanks a lot Stuart..

:) :)
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