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Bunuel

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01 Feb 2024, 00:29

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abhinav1239

Bunuel

Official Solution:

If Ben does not win the championship, the probability of Mike winning is \(\frac{1}{4}\), while the probability of Rob winning is \(\frac{1}{3}\). Given that the probability of Ben winning is \(\frac{1}{7}\), what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?

A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)

This problem involves the concept of conditional probability. It is important to understand that the direct probabilities of Mike and Rob winning the championship are not provided. Instead, we are given the probabilities of Mike and Rob winning the championship, only under the condition that Ben does not win. This means that the probabilities of Mike and Rob winning (\(\frac{1}{4}\) and \(\frac{1}{3}\), respectively) are relevant only if we consider the scenario where Ben does not win the championship.

The probability of Ben losing is \(1 - \frac{1}{7} = \frac{6}{7}\). Out of these \(\frac{6}{7}\) cases, the probability of Mike winning is \(\frac{1}{4}\) and the probability of Rob winning is \(\frac{1}{3}\). So the probability of either Mike or Rob winning the championship is \(P = \frac{6}{7}(\frac{1}{4} + \frac{1}{3}) = \frac{1}{2}\).

Alternatively, consider the following:

Take 84 championship cases (84 is chosen as it is the least common multiple of 3, 4, and 7).

Out of these 84 cases, Ben will lose in \(\frac{6}{7} *84 = 72\). When Ben loses, Mike would win in \(72*\frac{1}{4} = 18\) cases (which is one-fourth of the cases when Ben loses), and Rob would win in \(72*\frac{1}{3} = 24\) cases. Therefore, the probability of either Mike or Rob winning the championship is given by \(P = \frac{18 + 24}{84} = \frac{1}{2}\).

Answer: C

Bunuel a baisc conceptual doubt. For the case where Mike wins, could you please clarify why you don't need to multiply the probability of Rob not winning e.g. 1/6*1/4*2/3

The 1/4 probability for Mike to win is conditioned only on the event of Ben not winning. So, that probability for Mike to win, in the case of Ben losing, already accounts for the cases of Rob losing.

CoffeeNCream

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27 Feb 2024, 23:52

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Yes, I also mis interpreted the question as they were all competing at the same time.

Bunuel, just a question, IF the question had said they were all competing at the same time, would my line of thinking be correct?

(6/7 * 1/4 * 2/3) + (6/7 * 3/4 *1/3)?

Thanks!

Bunuel

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28 Feb 2024, 00:02

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CoffeeNCream

Whether the participants compete at the same time or not is not specified in the question and is irrelevant. Regardless of whether they compete together or not, the probability of Mike and Rob winning is given for the case when Ben loses.

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The solution relies on the law of total probability.

we are given
P(M | not B) = 1/4
P(R | not B) = 1/3
P(not B) = 6/7

The answer says

P(M) = P(M | not B) * P(not B) = 1/4 * 6/7 = 3/14
P(R) = P(R | not B) * P(not B) = 1/3 * 6/7 = 4/14
P(M) + P(R) = 7/14

Now why does P(M) = P(M | not B) * P(not B)? This is because the law of total probability says:

P(M) = P(M | not B) * P(not B) + P(M | B) * P(B)

In this case, P(M | B) * P(B) = 0 because there cannot be a case where M wins in even that B wins. So this becomes

P(M) = P(M | not B) * P(not B).

Bunuel

If Ben does not win the championship, the probability of Mike winning is \(\frac{1}{4}\), while the probability of Rob winning is \(\frac{1}{3}\). Given that the probability of Ben winning is \(\frac{1}{7}\), what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?

A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)

kll08

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14 Dec 2024, 23:10

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Bunuel will it be possible for you to add more questions on the conditional probability to master this concept?

Bunuel

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15 Dec 2024, 01:54

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kll08

Bunuel will it be possible for you to add more questions on the conditional probability to master this concept?

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03 Jan 2025, 23:28

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Bunuel

I struggle with probability. Can you please share some resources that will help me solve such questions?

Bunuel

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04 Jan 2025, 02:45

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arnavghatage

Bunuel

I struggle with probability. Can you please share some resources that will help me solve such questions?

22. Probability

For more:
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Hope it helps.

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29 Jan 2025, 08:38

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I like the solution - it’s helpful.

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20 Jul 2025, 07:09

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I like the solution - it’s helpful.

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31 Aug 2025, 06:39

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I like the solution - it’s helpful. I really liked alternative way to solve

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21 Sep 2025, 11:10

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Hi KarishmaB Bunuel how do we know here that we are not asked direct probabilities of Mike or Ross winning but conditional probabilities as the ques stem ''what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?'' isnt pretty clear and from this part of the ques, it seems that they are asking us to find direct probabilities of Mike & Ross from the conditional probabilities that they have given.

Bunuel

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21 Sep 2025, 11:15

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Natansha

The stem itself says: “If Ben does not win the championship, the probability of Mike winning is 1/4, while the probability of Rob winning is 1/3.”

That wording makes it clear these are conditional probabilities (only valid when Ben doesn’t win), not direct probabilities. If they were direct, the stem would simply say “The probability of Mike winning is 1/4” without mentioning Ben.