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Probability Problem

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by gmat2011 » Tue Jul 05, 2011 12:04 am
Hi all ,

I have come across the below problem in Probability and need your help to solve the same .

Problem
-------

Kamal and Monica appeared for an interview for two vacancies . The probability of Kamal's selection is 1/3 and that of Monica's rejection is 4/5 . Find the probability that only one of them will be selected .

Solution suggested by the book :
------------------------------

Let K be the event that Kamal will be selected and M the event that Monica will be selected .
Then P(K) = 1/3 and P(M^c) = 4/5
Therefore P(M) = 1 - P(M^c) = 1-4/5 = 1/5
and P(K^c) = 1-1/3 = 2/3

Only one of them will be selected if either i.) Kamal is selected and Monica is not selected or ii.) Monica is selected and Kamal is not

In case i.) Probability = 1/3 * 4/5 = 4/15
In case ii.) Probability = 2/3 * 1/5 = 2/15

Hence the required probability , that one of them will be selected is
4/15 + 2/15 = 6/15= 2/5

My approach
-----------

Required probability = P(K) + P(M) -P(K).P(M)
= 1/3 + 1/5 - 1/15
= 7/15

Could someone tell me where I am making a mistake .
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by ntamhane » Tue Jul 05, 2011 4:13 am
Probability of Kamal getting selected is 1/3
Probability of kamal getting rejected is 1-1/3 = 2/3

Probability of monica getting rejected is 4/5
Probability of kamal getting selected is 1/5

now our challenge is to find only one will be selected. That is possible only when
kamal is selected and monica is rejected or monica is selected and kamal is rejected.

hence [spoiler]1/3*4/5 + 2/3*1/5 = 2/5[/spoiler]
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by Frankenstein » Tue Jul 05, 2011 4:57 am
gmat2011 wrote: My approach
-----------

Required probability = P(K) + P(M) -P(K).P(M)
= 1/3 + 1/5 - 1/15
= 7/15

Could someone tell me where I am making a mistake .
Hi,
You are calculating P(K or M) by using that formula. So, you are including the probability of selecting both as well.
P(K or M) = P(K) + P(M) -P(K and M)

But, we need P(only K or only M). So, for this we need to calculate P(K or M) - P(K and M).
So, you will get 7/15-1/15 = 6/15
Cheers!

Things are not what they appear to be... nor are they otherwise
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by bharathram » Tue Jul 05, 2011 10:25 am
But, we need P(only K or only M). So, for this we need to calculate P(K or M) - P(K and M).
So, you will get 7/15-1/15 = 6/15
Sorry, but I don't understand this step. Can you explain this formula P(K or M) - P(K and M) better with the values?
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by Jim@Knewton » Tue Jul 05, 2011 5:44 pm
Pr| K selected = 1/3
Pr| K not selected = 2/3
Pr| M selected = 1/5
Pr| M not selected = 4/5

Quickest method (Recommended):
Pr| Only one selected = (K selected)*(M not selected) + (M selected)*(K not selected)
=> Pr| Only one selected = (1/3)* (4/5) + (1/5)*(2/3) = [spoiler]6/15 = 2/5[/spoiler]

Alternative method (Stylish!): ('Only K or only M' is the same as 'only one')
Pr| Only K or only M = Pr| K selected + Pr| M selected - Pr| both selected
=> Pr| Only one selected = 1/3 + 1/5 - (1/3)*(1/5) - (1/3)*(1/5)
=> Pr| Only one selected = [spoiler]8/15 - 2/15 = 6/15 = 2/5
[/spoiler]
Gmat2011 - you set it up very well but missed subtracting the additional "(1/3)*(1/5)"
The reason we need to subtract "(1/3)*(1/5)" twice is because the selection (or rejection) of K and M are two entirely independent events.
=>If you subtract "(1/3)*(1/5)" only once, then it would be adjusted in either Pr| K or Pr| M but not both.
Another way of stating this is that, "(1/3)*(1/5)" possibility is intrinsically present in both: Pr| K selected = 1/3 and Pr| M selected = 1/5 and hence needs to be subtracted from each of them.
Hope this helps! :-)
Best, Jim
Please "thank" this post if it helps you!
https://www.knewton.com/people/jims
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