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The table below shows [#permalink] New post 23 Dec 2016, 15:08
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The table below shows the distribution of a group of 40 college students by gender and class.


Attachment:
#GREexcercise The table below shows the distribution.jpg
#GREexcercise The table below shows the distribution.jpg [ 22.13 KiB | Viewed 3557 times ]



If one student is randomly selected from this group, find the probability that the student chosen is
(a) not a junior
(b) a female or a sophomore
(c) a male sophomore or a female senior

[Reveal] Spoiler: OA
\((a) \frac{21}{40} (b) \frac{7}{10} (c) \frac{9}{40}\)


Math Review
Question: 13
Page: 297
Difficulty: medium

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Re: The table below shows [#permalink] New post 26 Mar 2018, 22:03
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Let P(Female) = 22/40
Let P(Sophomore) = 16/40
then
P(female U sophomore) = 22/40 + 16/40 - (10/40) = 28/40
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Re: The table below shows [#permalink] New post 09 Apr 2018, 19:35
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HarveyKlaus wrote:
The table shows the distribution of a group of 40 college students by gender and class (in the pic), If one student is randomly selected from this group, find the probability that the student chosen is

(i) a male sophomore or a female senior
(ii) a female or a sophomore

My approach is:

(i) a male sophomore or a female senior
Sol: I divide the prompt in two section. the first is a male sophomore => 6 / 40 and the second part is a female senior => 3 / 40. Since its OR in-between these two categories, we add both after simplification => 3/20 + 3/40 ==> 9 / 40. And the answer is correct.

However, when I apply the same logic to the second prompt (ii) " a female or a sophomore "
Sol: First part --> Since there are overall 22 females in the whole population, the prob. of selecting a female would be 22 / 40 or 11 / 20. Secondly --> Since there are 16 sophomores in the whole population, the prob. of selecting a sophomore would be 16 / 40 or 2 / 5. Since there is OR in-between these two main categories, we add 11 / 20 + 2 / 5 ==> 19/20. BUT this is not the correct answer. (The correct answer is 28 / 40)


Problem: In the first prompt, my logic and the way I approach this problem works however, in the second prompt, it does not. What am I missing?

Thanks for your help!
H.



What you are missing is that there are female students who are sophomores and you are counting them twice while calculating your probability. So you can subtract the probability of females who are sophomores. In the first case, there is no intersection of a male sophomore and female senior i.e there won't be any male sophomore who would be a female senior as well.
Hope it helps
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Re: The table below shows [#permalink] New post 04 May 2018, 00:25
my answer was wrong too for the second problem. thanks @sam_ridhi and @novice07 for your explanation.
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Re: The table below shows [#permalink] New post 17 May 2018, 22:03
so for the second section, we should basically consider it as females or male sophomores?
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Re: The table below shows [#permalink] New post 02 Jun 2018, 03:55
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P(A or B) = P(A) + P(B) - P(A and B)

(i)

P(Male sophomer) = 6/40
P(Female senior) = 3/40
P(Male sophomer and female senior) = Not possible = 0

So, P(Male sophomer or Female senior)= P(Male sophomer) + P(Female senior) - P(Male sophomer and female senior)
= 6/40 + 3/40 - 0
= 9/40


(ii)

P(Female) = 22/40
P(Sophomore) = 16/40
P(Female and sophomer) = 10/40

So, P(female or sophomore) = P(Female) + P(Sophomer) - P(Female and sophomer)
= 22/40 + 16/40 - (10/40)
= 28/40
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Re: The table below shows [#permalink] New post 05 Jun 2018, 00:52
Got knocked out! Thanks a lot for this
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Re: The table below shows the distribution of a group of 40 coll [#permalink] New post 12 Jun 2019, 06:31
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Carcass wrote:
The table below shows the distribution of a group of 40 college students by gender and class.

Attachment:
#GREexcercise The table below shows the distribution.jpg


If one student is randomly selected from this group, find the probability that the student chosen is
(a) not a junior
(b) a female or a sophomore
(c) a male sophomore or a female senior

[Reveal] Spoiler: OA
\((a) \frac{21}{40} (b) \frac{7}{10} (c) \frac{9}{40}\)


Math Review
Question: 13
Page: 297
Difficulty: medium


Explanation::

A) Total number of students = 40

and total number of juniors = 19

Hence the probability of not selecting a junior = \(\frac{19}{40}\)

B) Total number of female students = 22

Total number of sophomore = 16

here we have to use the OR probability ( event A = female and event B = sophomore)

i.e P(A or B) = P(A) + P(B) - P( A & B)

or P(female or a sophomore) = \(\frac{22}{40} + \frac{16}{40}- \frac{10}{40} = \frac{28}{40} = \frac{7}{10}\)
since the two events are dependent, so P(A & B) = \(\frac{10}{40}\) as the sophomore is counted twice (male and female)

C) Total number of male sophomore = 6

Total number of female senior = 3

Hence the probability ( using OR probability)of selecting a male sophomore or a female senior = \(\frac{6}{40} + \frac{3}{40} - 0 = \frac{9}{40}\)( since the two events are independent hence P (male sophomore & a female senior) = 0
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Re: The table below shows the distribution of a group of 40 coll [#permalink] New post 25 Oct 2019, 11:01
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Re: The table below shows [#permalink] New post 29 Jun 2020, 18:38
The number of people is 40 and with this, it is possible to obtain the probabilities with this value.
Re: The table below shows   [#permalink] 29 Jun 2020, 18:38
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