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Probability

by ash4gmat » Thu May 05, 2016 6:21 am
A high school gym teacher randomly selects a group of two players from three students to demonstrate a basketball drill during class. The three students consists of two girls, Andrea and Marta, and one boy, Davi.

Let A be the event that the first player the coach chooses is a girl and B be the event that the second player is a boy.
What is P(A or B),the probability that the coach chooses a girl first or a boy second?

I am confused with P(A and B): Does that not mean probability of selecting a boy or Girl which is equal to zero?
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by regor60 » Thu May 05, 2016 6:45 am
There are 6 different ways of picking two people:

Andrea and Marta

Andrea and then Davi

Marta and then Davi

Marta and then Andrea

Davi and then Andrea

Davi and then Marta


By inspection you can see that only the last two scenarios don't satisfy the question. Those last two represent 2 of 6 possible ways of picking two people, or 1/3.

Therefore, the probability of the OR scenario described in the question is:

1 - 1/3 = 2/3

The AND scenario means that BOTH events have to happen, meaning in this example since there are two ways for this to take place:

Andrea and Davi

Marta and Davi

Which equal 1/3 of the possible ways of picking two people
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by ash4gmat » Thu May 05, 2016 7:14 am
Regor with lesser variables it is easy to list down and calculate.With Higher values what is the approach?
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by jswesth » Fri May 06, 2016 1:13 pm
Why wouldn't this be 2/3 + 1/6 or 5/6 probability? At the beginning of the selection process, there are 2 girls and 1 boy to choose from meaning that the probability of selecting a girl would be 2/3. Because it's asking for "OR", we have to add the second probability that a boy will be picked second. The odds that the boy isn't chosen first is 1/3 and the odds that he is chosen second is 1/2. So the odds that he chosen exactly in the second spot is 1/6, right?

So adding 2/3 + 1/6 = 5/6? Where am I wrong here?
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by ash4gmat » Fri May 06, 2016 11:33 pm
Experts any other approach apart from listing down.
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by regor60 » Mon May 09, 2016 7:44 am
jswesth wrote:Why wouldn't this be 2/3 + 1/6 or 5/6 probability? At the beginning of the selection process, there are 2 girls and 1 boy to choose from meaning that the probability of selecting a girl would be 2/3. Because it's asking for "OR", we have to add the second probability that a boy will be picked second. The odds that the boy isn't chosen first is 1/3 and the odds that he is chosen second is 1/2. So the odds that he chosen exactly in the second spot is 1/6, right?

So adding 2/3 + 1/6 = 5/6? Where am I wrong here?
The rules for OR probability are that you add the individual probabilities as if they were mutually exclusive, then subtract the degree to which they are NOT actually mutually exclusive.

In other words, by simply adding the probabilities, you are also double counting an event that appears in both individual probabilities.

You can see from the list of selections in my response above, that there are four groups where a woman is selected first, hence the 2/3 probability. However, 2 of those include the boy second, which independently is 1/3 probability.

So P(A OR B) = P(A) + P(B) -P(A AND B) = 2/3 +1/3 -1/3 =2/3
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by regor60 » Mon May 09, 2016 7:53 am
jswesth wrote:. The odds that the boy isn't chosen first is 1/3 and the odds that he is chosen second is 1/2. So the odds that he chosen exactly in the second spot is 1/6, right?
No, the probability he isn't chosen first is 2/3, since 2 out of 3 people are women.

So, the probability that he is chosen second is 2/3 * 1/2 = 1/3
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by 800_or_bust » Mon May 09, 2016 11:46 am
jswesth wrote:Why wouldn't this be 2/3 + 1/6 or 5/6 probability? At the beginning of the selection process, there are 2 girls and 1 boy to choose from meaning that the probability of selecting a girl would be 2/3. Because it's asking for "OR", we have to add the second probability that a boy will be picked second. The odds that the boy isn't chosen first is 1/3 and the odds that he is chosen second is 1/2. So the odds that he chosen exactly in the second spot is 1/6, right?

So adding 2/3 + 1/6 = 5/6? Where am I wrong here?
The odds that Davi is selected exactly second are 1/3, not 1/6. Your error is that the probability that he isn't selected first is 2/3, not 1/3. Therefore, 2/3 x 1/2 = 1/3. Remember, if we were selecting three people, the probability of Davi being selected would have to be one. The odds that he would be selected first, second or third would each be 1/3.

With that said, it's obvious that 2/3 + 1/3 = 1 is not the correct answer. That would imply there is a 100% chance of a girl being selected first and Davi being selected second. Clearly, this is not true.
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by 800_or_bust » Mon May 09, 2016 11:54 am
regor60 wrote:
jswesth wrote:Why wouldn't this be 2/3 + 1/6 or 5/6 probability? At the beginning of the selection process, there are 2 girls and 1 boy to choose from meaning that the probability of selecting a girl would be 2/3. Because it's asking for "OR", we have to add the second probability that a boy will be picked second. The odds that the boy isn't chosen first is 1/3 and the odds that he is chosen second is 1/2. So the odds that he chosen exactly in the second spot is 1/6, right?

So adding 2/3 + 1/6 = 5/6? Where am I wrong here?
The rules for OR probability are that you add the individual probabilities as if they were mutually exclusive, then subtract the degree to which they are NOT actually mutually exclusive.

In other words, by simply adding the probabilities, you are also double counting an event that appears in both individual probabilities.

You can see from the list of selections in my response above, that there are four groups where a woman is selected first, hence the 2/3 probability. However, 2 of those include the boy second, which independently is 1/3 probability.

So P(A OR B) = P(A) + P(B) -P(A AND B) = 2/3 +1/3 -1/3 =2/3
Thanks for the explanation. And the earlier approach of listing out all of the possibilities was helpful. It's clear now why the overall probability of A or B occurring is totally dependent on P(A), because for P(B) to occur, P(A) must first occur! I was a little confused at first, because at first glance you would expect P(A or B) to be greater than P(A), since there are two possibilities instead of one. But the reality is there's really only one event that matters, since event two (i.e. P(B)) can only occur if the first event occurs.
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by 800_or_bust » Mon May 09, 2016 11:58 am
regor60 wrote:There are 6 different ways of picking two people:

Andrea and Marta

Andrea and then Davi

Marta and then Davi

Marta and then Andrea

Davi and then Andrea

Davi and then Marta


By inspection you can see that only the last two scenarios don't satisfy the question. Those last two represent 2 of 6 possible ways of picking two people, or 1/3.

Therefore, the probability of the OR scenario described in the question is:

1 - 1/3 = 2/3

The AND scenario means that BOTH events have to happen, meaning in this example since there are two ways for this to take place:

Andrea and Davi

Marta and Davi

Which equal 1/3 of the possible ways of picking two people
Just to add one comment. The case P(A & B) can also be calculated as the product P(A) x P(B|A) = 2/3 x 1/2 = 1/3. Remember the notation P(B|A) is the probability of B occurring, given that A has occurred. In this case, the probability that a boy is selected second, given that a girl has been selected first. Here, P(B|A) = 1/2. That is, if a girl is selected first, the probability of selecting a boy second is 1/2.
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by Matt@VeritasPrep » Wed May 11, 2016 11:42 pm
Let's provide two approaches. First, the technical, painful way that's been discussed.

"Or" in the mathematical sense means "at least one of", so "A or B" means "at least one of A and B", or "just A or just B or both".

We could find the answer by finding

P(A) + P(B) - P(A and B)

We have to subtract P(A and B) once because it's counted TWICE - once in P(A) and once in P(B) - and we only want to count it ONCE.

P(A) is easy: 2/3.

Finding P(B) is a little funky. There's a (1/3) chance we pick Davi first, in which case we can't pick him second. There's a (2/3) chance we DON'T pick him first, then a (1/2) chance we pick him second. Either of these scenarios is in play, and they're mutually exclusive (only one can happen), so P(B) = (1/3)*0 + (2/3)*(1/2), or (1/3).

P(A and B) = (2/3)*(1/2) = (1/3).

Then we've got

(2/3) + (1/3) - (1/3), or (2/3).
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by Matt@VeritasPrep » Wed May 11, 2016 11:44 pm
Now let's do it the easy way.

We want girl first or boy second. That means the only scenario we DON'T want is boy first. The probability of that is 1/3. We want everything BUT this: 1 - 1/3, or 2/3, and we're done!
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by Matt@VeritasPrep » Wed May 11, 2016 11:45 pm
One more way:

Boy second is a subset of girl first - that is, to pick Davi second, you MUST have picked a girl first anyway. So "boy second" is redundant, and you only need to find the probability of a girl first, which is 2/3.
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