Could please somebody help? I can’t get the correct answer.
There are 8 magazines lying on a table; 4 are fashion magazines and the other 4 are sports magazines. If 3 magazines are to be selected at random from 8 magazines, what is the probability that at least one of the fashion magazines will be selected?
a) 1/2
b) 2/3
c) 32/35
d) 11/12
e) 13/14 - correct
The way I am solving it is: Probability (fashion >=1) = Probability (fashion 1) or Probability (fashion 1and 2) + Probability (fashion 1 and 2 and 3)=4/8+4/8*3/7+4/8*3/7*2/6 = 11/14, but it is wrong.
Thanks!
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- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
Your approach is incorrect for two distinct reasons. When you read these reasons, remember that your method assumes that the ORDER of the selections matters - this is always the case when you multiply 'consecutive' probabilities.
(1) You've counted certain cases more than once:
- Your first probability of 4/8 covers ALL cases in which the FIRST magazine is a fashion magazine. That includes cases in which the first and second magazines are both fashion magazines - and the first and third, and ALL of them.
(2) You left out certain cases:
- You didn't treat the cases in which magazine 1 ISN'T a fashion magazine, but one or the other (or both) of the other 2 magazines are.
If you want to use your approach, here's how you have to do it:
(just #1) + (just #2) + (just #3) + (#1 and #2 only) + (#1 and #3 only) + (#2 and #3 only) + (all three)
Of the seven (!) expressions above, the ONLY one you've calculated correctly is (all three). For each of the other six, you have to include the probablilities that the other magazines are not fashion magazines in your multiplications. For instance, the quantity (just #1) is 4/8 * 4/7 * 3/6. You can work out the other five quantities (you already have the right answer for the last one) and add them together; you should get 13/14.
--
Of course, all the above is one big moot point, because the poster below you has correctly pointed out the easy approach to this problem: calculate the value of the OPPOSITE event (i.e., none of the magazines are fashion) and subtract from 1.
If you don't like combinations, you can do this with your method:
Probability that NONE of the magazines is a fashion magazine = probability that all of them are sports magazines
= 4/8 * 3/7 * 2/6 = 1/14
so
your probability = 1 - 1/14 = 13/14
(1) You've counted certain cases more than once:
- Your first probability of 4/8 covers ALL cases in which the FIRST magazine is a fashion magazine. That includes cases in which the first and second magazines are both fashion magazines - and the first and third, and ALL of them.
(2) You left out certain cases:
- You didn't treat the cases in which magazine 1 ISN'T a fashion magazine, but one or the other (or both) of the other 2 magazines are.
If you want to use your approach, here's how you have to do it:
(just #1) + (just #2) + (just #3) + (#1 and #2 only) + (#1 and #3 only) + (#2 and #3 only) + (all three)
Of the seven (!) expressions above, the ONLY one you've calculated correctly is (all three). For each of the other six, you have to include the probablilities that the other magazines are not fashion magazines in your multiplications. For instance, the quantity (just #1) is 4/8 * 4/7 * 3/6. You can work out the other five quantities (you already have the right answer for the last one) and add them together; you should get 13/14.
--
Of course, all the above is one big moot point, because the poster below you has correctly pointed out the easy approach to this problem: calculate the value of the OPPOSITE event (i.e., none of the magazines are fashion) and subtract from 1.
If you don't like combinations, you can do this with your method:
Probability that NONE of the magazines is a fashion magazine = probability that all of them are sports magazines
= 4/8 * 3/7 * 2/6 = 1/14
so
your probability = 1 - 1/14 = 13/14
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