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Re: Need help in probability of 2 or more events [#permalink]
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skovinsky wrote:
jovic1104 wrote:
Hi,

Can anyone advise if my below solution is correct for the given probability question.

In a certain college, 100 out of 500 males are above 16 years old and 80 out of 400 females are above 16 years of age. What is the probability that a student selected for a particular grant is a male or is above 16 years of age?

My Solution:

P(Male) = 500/900
P(Female above 16 yrs. of age) = 400/900

P((Male) or (Female above 16 yrs. of age)) = P(Male) + P(Female above 16 yrs. of age) – P(Male and Female above 16 yrs. of age)

= 500/900 + 400/900 – 1/900
= 899/900


Appreciate for the feedback.


Hi!

Please provide the source and answer choices for all of your questions. Spoilers with the correct answers aren't very useful without the choices! Further, the answers often provide valuable hints as to how you should proceed.

In any case:

Probability = (# of desired outcomes)/(total # of possibilities)

Here, there are 900 students in total, so that will be our denominator. The numerator will be the # of students that match our requirements: male (any) or female that's over 16 years old. Accordingly, our equation becomes:

Probability = (# of males + # of females over 16)/(total # of students)

As you noted, there are 500 males. When you did your calculation, however, you set the number of females over 16 to 400; that's the total number of females, not the ones that we want to select. According to the question stem, there are only 80 females over 16. So, our numerator will be 500 + 80 = 580.

Consequently, the correct answer is:

580/900 = 58/90 = 29/45

As an important aside, the question isn't properly written. On the GMAT, the question would certainly let us know that the selection of a grant recipient is random - without that explicitly stated, there is no correct answer to the question. So, to be a proper question, it should read:

Quote:
In a certain college, 100 out of the 500 males are above 16 years old and 80 out of the 400 females are above 16 years of age. If one student is selected at random to receive a grant, what is the probability that the student selected is male or is above 16 years of age?


Wait a second here... doesn't the original prompt ask for the probability of a male student OR a student of any gender with an age over 16?

Let's solve each one and add them.

Probability of a male student:
500 males / 900 students overall = \(\frac{5}{9}\)

Probability of a student being over 16:
((100 males over 16) + (80 females over 16) ) / (900 students overall)

\(\frac{(100+80)}{(900)}\)

\(\frac{180}{900} =\frac{1}{5}\)

Since we're going for a combined probability (male OR over 16), we must add the two probabilities together.

Probability of being male + probability of being over 16 = total probability
\(\frac{5}{9} + \frac{1}{5} = \frac{34}{45}\)
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Re: Need help in probability of 2 or more events [#permalink]
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YourDreamTheater wrote:
Wait a second here... doesn't the original prompt ask for the probability of a male student OR a student of any gender with an age over 16?


It does, but since the two aren't mutually exclusive you can't simply add them together - by doing so, you're double counting all the males over 16.

If you want to tackle it the way you've proposed, you need to use the formula in the original post:

Prob(event A or B) = Prob(A) + Prob(B) - Prob(A and B)

Letting A = male and B = over 16:

Prob(A) = 500/900
Prob(B) = 180/900
Prob(A&B) = 100/900

Prob(A or B) = 500/900 + 180/900 - 100/900 = 580/900 = 29/45

Of course, that's a lot more work than just adding the two exclusive events that we do want (which is how I originally solved).
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Re: Need help in probability of 2 or more events [#permalink]
skovinsky wrote:
jovic1104 wrote:
Hi,

Can anyone advise if my below solution is correct for the given probability question.

In a certain college, 100 out of 500 males are above 16 years old and 80 out of 400 females are above 16 years of age. What is the probability that a student selected for a particular grant is a male or is above 16 years of age?

My Solution:

P(Male) = 500/900
P(Female above 16 yrs. of age) = 400/900

P((Male) or (Female above 16 yrs. of age)) = P(Male) + P(Female above 16 yrs. of age) – P(Male and Female above 16 yrs. of age)

= 500/900 + 400/900 – 1/900
= 899/900


Appreciate for the feedback.


Hi!

Please provide the source and answer choices for all of your questions. Spoilers with the correct answers aren't very useful without the choices! Further, the answers often provide valuable hints as to how you should proceed.

In any case:

Probability = (# of desired outcomes)/(total # of possibilities)

Here, there are 900 students in total, so that will be our denominator. The numerator will be the # of students that match our requirements: male (any) or female that's over 16 years old. Accordingly, our equation becomes:

Probability = (# of males + # of females over 16)/(total # of students)

As you noted, there are 500 males. When you did your calculation, however, you set the number of females over 16 to 400; that's the total number of females, not the ones that we want to select. According to the question stem, there are only 80 females over 16. So, our numerator will be 500 + 80 = 580.

Consequently, the correct answer is:

580/900 = 58/90 = 29/45

As an important aside, the question isn't properly written. On the GMAT, the question would certainly let us know that the selection of a grant recipient is random - without that explicitly stated, there is no correct answer to the question. So, to be a proper question, it should read:

Quote:
In a certain college, 100 out of the 500 males are above 16 years old and 80 out of the 400 females are above 16 years of age. If one student is selected at random to receive a grant, what is the probability that the student selected is male or is above 16 years of age?



Hi, thanks for the quick reply.

My understanding with the question is that since "male" covers all males, the "16 years of age" must be apply to females. So the probability formula must be

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

But im not really sure if this is correct. Guide me please. Thanks.
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Re: Need help in probability of 2 or more events [#permalink]
skovinsky wrote:
YourDreamTheater wrote:
Wait a second here... doesn't the original prompt ask for the probability of a male student OR a student of any gender with an age over 16?


It does, but since the two aren't mutually exclusive you can't simply add them together - by doing so, you're double counting all the males over 16.

If you want to tackle it the way you've proposed, you need to use the formula in the original post:

Prob(event A or B) = Prob(A) + Prob(B) - Prob(A and B)

Letting A = male and B = over 16:

Prob(A) = 500/900
Prob(B) = 180/900
Prob(A&B) = 100/900

Prob(A or B) = 500/900 + 180/900 - 100/900 = 580/900 = 29/45

Of course, that's a lot more work than just adding the two exclusive events that we do want (which is how I originally solved).


Quick question please. How did you get the Prob(A&B) = 100/900. Is it because of 100 males are above 16 yrs. of age?
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Re: Need help in probability of 2 or more events [#permalink]
skovinsky wrote:
YourDreamTheater wrote:
Wait a second here... doesn't the original prompt ask for the probability of a male student OR a student of any gender with an age over 16?


It does, but since the two aren't mutually exclusive you can't simply add them together - by doing so, you're double counting all the males over 16.

If you want to tackle it the way you've proposed, you need to use the formula in the original post:

Prob(event A or B) = Prob(A) + Prob(B) - Prob(A and B)

Letting A = male and B = over 16:

Prob(A) = 500/900
Prob(B) = 180/900
Prob(A&B) = 100/900

Prob(A or B) = 500/900 + 180/900 - 100/900 = 580/900 = 29/45

Of course, that's a lot more work than just adding the two exclusive events that we do want (which is how I originally solved).


Whoops! Thanks so much for cleaning up my obvious mistake!!!
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Re: Need help in probability of 2 or more events [#permalink]
jovic1104 wrote:
skovinsky wrote:
YourDreamTheater wrote:
Wait a second here... doesn't the original prompt ask for the probability of a male student OR a student of any gender with an age over 16?


It does, but since the two aren't mutually exclusive you can't simply add them together - by doing so, you're double counting all the males over 16.

If you want to tackle it the way you've proposed, you need to use the formula in the original post:

Prob(event A or B) = Prob(A) + Prob(B) - Prob(A and B)

Letting A = male and B = over 16:

Prob(A) = 500/900
Prob(B) = 180/900
Prob(A&B) = 100/900

Prob(A or B) = 500/900 + 180/900 - 100/900 = 580/900 = 29/45

Of course, that's a lot more work than just adding the two exclusive events that we do want (which is how I originally solved).


Quick question please. How did you get the Prob(A&B) = 100/900. Is it because of 100 males are above 16 yrs. of age?


Ok. I got why Prob(A&B) is 100/900. I have multiplied both numeration of A & B and reduced to lower fraction.

thanks.
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Re: Need help in probability of 2 or more events [#permalink]
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jovic1104 wrote:

Quick question please. How did you get the Prob(A&B) = 100/900. Is it because of 100 males are above 16 yrs. of age?


Correct!

Another way to think of this problem is as an overlapping sets question. We have a big group (all students) who have two different characteristics (maleness and over 16). The question is asking us what fraction of all students are male and/or over 16.

For 2 characteristics, the formula is:

Total # = (# with char 1) + (# with char 2) + (# with neither char) - (# with both char)

In this problem, we have 900 students total, 500 males, 180 over 16s and 320 who are neither male nor over 16 (i.e. the 16 and under females).

900 = 500 + 180 + 320 - both

900 = 1000 - both

both = 100

So, the total number of students who are male and/or over 16 is:

500 + 180 - 100 = 580

and the fraction of all students who are male and/or over 16 is:

580/900
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Re: Need help in probability of 2 or more events [#permalink]
skovinsky wrote:
jovic1104 wrote:

Quick question please. How did you get the Prob(A&B) = 100/900. Is it because of 100 males are above 16 yrs. of age?


Correct!

Another way to think of this problem is as an overlapping sets question. We have a big group (all students) who have two different characteristics (maleness and over 16). The question is asking us what fraction of all students are male and/or over 16.

For 2 characteristics, the formula is:

Total # = (# with char 1) + (# with char 2) + (# with neither char) - (# with both char)

In this problem, we have 900 students total, 500 males, 180 over 16s and 320 who are neither male nor over 16 (i.e. the 16 and under females).

900 = 500 + 180 + 320 - both

900 = 1000 - both

both = 100

So, the total number of students who are male and/or over 16 is:

500 + 180 - 100 = 580

and the fraction of all students who are male and/or over 16 is:

580/900



I appreciate well your explanation. Thanks.



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