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# If the probability that Mike can win a championship is 1/4,

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If the probability that Mike can win a championship is 1/4, [#permalink]  01 Apr 2007, 18:01
If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6. What is the probability that Mike or Rob win the championship but Ben don't?
A) 1/6
B)5/12
C) 7/12
D)5/6
E)1
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If you have OG 11th edition, open it at page no. 184, Q 231...it is almost same as the question given here but the only difference is: AND, not OR...I mean it is like this:

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6. What is the probability that Mike and Rob win the championship but Ben don't?

Then the solution according to OG is:

Prob of Ben not_winning = 1 - 1/6 = 5/6

Prob of Mike_AND_Rob_notBen = 1/4 * 1/3 * 5/6 = 5/72

But we have the question like this:

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6. What is the probability that Mike or Rob win the championship but Ben don't?

Prob of Ben not_winning = 1 - 1/6 = 5/6
Prob of Mike_OR_Rob = 1/4 + 1/3 = 7/12
Prob of Mike_OR_Rob_notBen = 7/12 * 5/6 = 35/72

Prob of Mike_OR_Rob = 1/4 + 1/3 = 7/12
Prob of Mike_OR_Rob_notBen = 7/12 - 1/6 = 5/12 But I don't know how we can do this???
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I think it is from Kaplan.

C 7/12 for me. Add Mike's and Rob's rate together.
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C for me too.

Although I did it differently...

(1/4) + (1/3) = 7/12 Mike OR Rob (so add)

prob of NOT Ben = 5/6

(7/12)*5/6 = 35/72 (slightly over 1/2)

the only ans choice that is slightly over 1/2 is C.

What am I doing wrong??

tennis_ball, why didn't you include p(not Ben)? Is this not a (AuB)nC situation???
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When I did it, I got it wrong. So this time around, I am wrong again. Actually I don't understand why you have to subtract Ben away. Any pointers?
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Summer3 wrote:
...

But we have the question like this:

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6. What is the probability that Mike or Rob win the championship but Ben don't?

Prob of Ben not_winning = 1 - 1/6 = 5/6
Prob of Mike_OR_Rob = 1/4 + 1/3 = 7/12
Prob of Mike_OR_Rob_notBen = 7/12 * 5/6 = 35/72

Prob of Mike_OR_Rob = 1/4 + 1/3 = 7/12
Prob of Mike_OR_Rob_notBen = 7/12 - 1/6 = 5/12 But I don't know how we can do this???

The reason why, I think, is because Mike, Rob and Ben are in the same competition. Their chances on winning are incompatible, because the event of Mike winning, Rob winning or Ben winning cannot happen at the same time...there can be only 1 winner. For this reason we need to use the "simple" OR formula: P(A OR B) = P(A) + P(B) although in this case we have 3 events thus: P(A OR B OR C) = P(A) + P(B) + P(C) as opposed to the "complex" OR formula for compatible events (can happen at the same time) which looks like this P(A OR B) = P(A) + P(B) - P(A AND B). (the reason why I show the "complex" OR is because the above setup used by Summer3: Prob of Mike_OR_Rob_notBen looks like it)

Anyway, back to the problem, we know that we have to use the "simple" OR with 3 events (A, B and C..which are Mike winning, Rob winning and Ben winning respectively), so let's fill it in and customise it (bold):

P(A OR B OR NOT C) = P(A) + P(B) - P(C)
= 1/4 + 1/3 - 1/6
= 7/12 - 2/12
= 5/12, so answer B indeed.

The reason why, when I customised the "simple" OR formula, I changed the second + sign into a -, is because of the following:

The probability of Mike, Rob and Ben winning are different. No one knows why, so let's assume it's a running championship and Ben is in a wheelchair, Rob has one artificial leg and Mike is the only person without a leg problem. In that case, it is logical to assume their respective probabilities for winning. So, in order to find out what the probability of winning is for Mike OR Rob but not Ben, we must subtract Ben's probability as...and this is crucial for understanding this...it is because of Ben's wheelchair condition that the others' probabilities are different. Every person's probability to win is effected by each other's presence.

I hope this explanation is correct, but I cannot guarantee it, so it really needs an expert inspection cause I'm just a Joe Blogg when it comes to math

P.s. credit to my girl as she is largely responsible for making me understand it
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