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Mike or Rob win but not Ben

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by sanju09 » Sat Feb 21, 2009 2:41 am
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 not Ben?

Produce your own response, links :)
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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Source: — Problem Solving |

Re: Mike or Rob win but not Ben

by bluementor » Mon Feb 23, 2009 5:01 am
sanju09 wrote: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 not Ben?

Produce your own response, links :)
Prob(Ben looses AND Mike wins) = (5/6)* (1/4)
Prob(Ben looses AND Mike wins) = (5/6)* (1/3)

Prob(Mike OR Rob wins AND Ben looses) = (5/6)*(1/4) + (5/6)*(1/3) = (5/6)*(1/3 + 1/4) =(5/6)*(7/12)=35/72
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Re: Mike or Rob win but not Ben

by sanju09 » Mon Feb 23, 2009 5:05 am
bluementor wrote:
sanju09 wrote: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 not Ben?

Produce your own response, links :)
Prob(Ben looses AND Mike wins) = (5/6)* (1/4)
Prob(Ben looses AND Mike wins) = (5/6)* (1/3)

Prob(Mike OR Rob wins AND Ben looses) = (5/6)*(1/4) + (5/6)*(1/3) = (5/6)*(1/3 + 1/4) =(5/6)*(7/12)=35/72
Well... a clue

winning a championship is a mutually exclusive event B-)
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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Re: Mike or Rob win but not Ben

by sureshbala » Mon Feb 23, 2009 5:19 am
sanju09 wrote: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 not Ben?

Produce your own response, links :)
Case I: Mike wins and both Rob and Ben looses

i.e 1/4 x 2/3 x 5/6

Case II: Rob wins and Mike and Ben looses

i.e 1/3 x 3/4 x 5/6

Hence required probability = 5/6( 1/6+1/4) = 25/72
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Re: Mike or Rob win but not Ben

by x2suresh » Mon Feb 23, 2009 10:10 am
sanju09 wrote: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 not Ben?

Produce your own response, links :)
=1/4+1/3
= 7/12
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by welcome » Mon Feb 23, 2009 12:09 pm
(M & noR & NOB) OR (NOM & R & NOB)

1/4 * 2/3 * 5/6 + 3/4 * 1/3 * 5/6 = 25/72 - ANSWER

What is OA?
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590
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by benjaminvw » Mon Feb 23, 2009 4:59 pm
Given that they're mutually exclusive I don't get why it isn't 1/3 + 1/4 = 7/12

That said, these probably questions are some of the hardest for me......
Benjaminvw
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by x2suresh » Mon Feb 23, 2009 5:46 pm
benjaminvw wrote:Given that they're mutually exclusive I don't get why it isn't 1/3 + 1/4 = 7/12

That said, these probably questions are some of the hardest for me......
oK one more person supporting my answer.

I believe we are correct.
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by krisraam » Sun Mar 01, 2009 9:53 am
Probability that Mike or Rob winning a championship

= 1- ( Mike and Rob both loosing the championship)

= 1 -(3/4*2/3)

= 1/2

Probablity that Ben loosing a championship = 5/6

Probability that Mike or Rob winning a championship and Ben loosing is

1/2*5/6 = 5/12

Whats OA

Thanks
raama
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by sanju09 » Mon Mar 02, 2009 2:24 am
krisraam wrote:Probability that Mike or Rob winning a championship

= 1- ( Mike and Rob both loosing the championship)

= 1 -(3/4*2/3)

= 1/2

Probablity that Ben loosing a championship = 5/6

Probability that Mike or Rob winning a championship and Ben loosing is

1/2*5/6 = 5/12

Whats OA

Thanks
raama
OA is seven upon twelve
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion

by krisraam » Mon Mar 02, 2009 8:35 am
sanju09 wrote:
krisraam wrote:Probability that Mike or Rob winning a championship

= 1- ( Mike and Rob both loosing the championship)

= 1 -(3/4*2/3)

= 1/2

Probablity that Ben loosing a championship = 5/6

Probability that Mike or Rob winning a championship and Ben loosing is

1/2*5/6 = 5/12

Whats OA

Thanks
raama
OA is seven upon twelve
I still think my answer is correct.

If all three are attending the same championship.

The probability of Mike or Rob winning the championship( That implies Ben loosing) is 1/4 + 1/3 = 7/12

If Mike, Rob, Ben are competing in different championships( Tennis,Golf,Swimming) and the numbers given are probabilities of winning in their respective championships.

Then the probability of Mike and Rob winning and Ben loosing is 5/12.

I assumed this way because we don't need the probabilty of Ben winning a championship if all are participating the same championship.

Some where in the chain it was mentioned they are mutually exclusive. How would they be mutually exclusive if all three participate in the same championship.

If Mike wins the championship, everyone else has to lose right??


Thanks
raama
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by benjaminvw » Mon Mar 02, 2009 12:35 pm
I'm confused by raama's statement and conclusion, he says it himself, "The probability of Mike or Rob winning the championship( That implies Ben loosing) is 1/4 + 1/3 = 7/12." That's what we're looking for.

Mutually exclusive means that by default if Mike or Rob wins then Ben loses. So we don't have to account for him. It's like when we find the probably of rolling a 2 on a six sided dice, it's just 1/6. We don't have to worry about the odds of rolling a 1, 3, 4, 5, or 6.

If I'm wrong, someone please tell me, but that's how I see it: 1/3 + 1/4
Benjaminvw
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by krisraam » Mon Mar 02, 2009 1:01 pm
benjaminvw wrote:I'm confused by raama's statement and conclusion, he says it himself, "The probability of Mike or Rob winning the championship( That implies Ben loosing) is 1/4 + 1/3 = 7/12." That's what we're looking for.

Mutually exclusive means that by default if Mike or Rob wins then Ben loses. So we don't have to account for him. It's like when we find the probably of rolling a 2 on a six sided dice, it's just 1/6. We don't have to worry about the odds of rolling a 1, 3, 4, 5, or 6.

If I'm wrong, someone please tell me, but that's how I see it: 1/3 + 1/4
My Bad. I misunderstood mutually exclusive events as events whose outcome doesn't affect the outcome of the next event. Like in case of tossing a coin 3 times. Each outcome is independent and doesn't depend on the previous outcome.

My reasoning was in those lines.

Assume that Mike is participating in Tennis tournament
Rob into Swimming, Ben into gymnastics.

Mike winning a the championship has no effect Rob winning the swimming championship....

That lead me to the answer 5/16

I overthinked the question and went too far.

Thanks
raama
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