Last visit was: 12 Jul 2024, 16:55 It is currently 12 Jul 2024, 16:55

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [2]
Given Kudos: 25479
Send PM
Retired Moderator
Joined: 07 Jan 2018
Posts: 739
Own Kudos [?]: 1394 [4]
Given Kudos: 93
Send PM
avatar
Director
Director
Joined: 09 Nov 2018
Posts: 505
Own Kudos [?]: 131 [0]
Given Kudos: 0
Send PM
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 451 [1]
Given Kudos: 0
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
1
Expert Reply
Carcass wrote:
A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)



OK... the two ways are

1) Cases when none are there..
Total ways = (7+3)(7+3-1)=10*9=90
ways when both are men, => 3*2=6
thus ways when atleast one is woman => 90-6=84
Thus probability that atleast one is woman = \(\frac{84}{90}=\frac{14}{15}\)

2) find each case when women are present
a) only one is present
choose one female = 7C1
choose one male = 3C1
total ways = 7C1*3C1=7*3=21
but the female can be at either of the two position => 21*2=42
b) both are women
first one can be any of 7 and next any of remaining 6, so 7*6=42

Total ways women are present = 42+42=84
probability = \(\frac{84}{90}=\frac{14}{15}\)


A
Retired Moderator
Joined: 07 Jan 2018
Posts: 739
Own Kudos [?]: 1394 [0]
Given Kudos: 93
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
1
AE wrote:
amorphous wrote:
we can solve this problem by a direct method which is more time consuming.
i.e finding the no of ways that at least one women is selected.


For understanding, could please explain that too.


At least 1 woman is selected if women and women is selected or women and men is selected or men and women is selected.

total number of possible outcomes is \(10 * 9 = 90\)

w&w = \(7*6 = 42\)
w&m = \(7*3 = 21\)
m&w = \(3*7 = 21\)
therefore the probability that at least 1 woman is selected is \(\frac{84}{90}\)which reduces to \(\frac{14}{15}\)
avatar
Intern
Intern
Joined: 26 Oct 2018
Posts: 8
Own Kudos [?]: 5 [0]
Given Kudos: 0
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
We can solve this using combination.
1 - (All Men Probability).

All men probability = 3C2 / 10C2 = 3/45 = 1/15

At least 1 Woman probability = 1-(1/15) = 14/15
Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [0]
Given Kudos: 25479
Send PM
A club has exactly 3 men and 7 women as members. If two memb [#permalink]
Expert Reply
A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11864 [0]
Given Kudos: 136
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
3
Carcass wrote:
A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)


We can solve this by using counting methods or by applying probability rules.
Let's use probability rules

We want P(selecting at least 1 woman)

When it comes to probability questions involving "at least," it's best to try using the complement.

That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(selecting at least 1 woman) = 1 - P(not selecting at least 1 woman)
What does it mean to not select at least 1 woman? It means selecting ZERO women.
So, we can write: P(selecting at least 1 woman) = 1 - P(selecting ZERO women)


P(selecting ZERO women) = P(selecting TWO men)
= P(selecting a man to be president AND selecting a man to be vice-president)
= P(selecting a man to be president) x P(selecting a man to be vice-president)
= 3/10 x 2/9
= 6/90
= 1/15
ASIDE: For the 1st selection, 3 of the 10 people are men.
Once we select a man for the 1st selection, there are 9 people remaining, and 2 of them are men.


So, P(selecting at least 1 woman) = 1 - 1/15
= 14/15

Answer: A

Cheers,
Brent
avatar
Intern
Intern
Joined: 02 Mar 2019
Posts: 45
Own Kudos [?]: 13 [0]
Given Kudos: 0
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
1
A very comprehensive explanation as always, Brent!
The answer should be 'A', I think you made a typo saying 'E'.
Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [0]
Given Kudos: 25479
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
Expert Reply
Thank you.

Fixed
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11864 [0]
Given Kudos: 136
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
LT2018 wrote:
A very comprehensive explanation as always, Brent!
The answer should be 'A', I think you made a typo saying 'E'.


Thanks LT2018!
I've edited my response.

Cheers,
Brent
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3017 [1]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
1
There are 3M and 7W

Prob. (1W for at-least one position) = 1 - (No W for both the positions)
So, we can select 2M from 3 in 3C2 ways.

Therefore,
Required Prob. = 1 - [3C2 / 10C2] = 14/15

Hence, option A
Manager
Manager
Joined: 11 Oct 2023
Posts: 69
Own Kudos [?]: 40 [1]
Given Kudos: 25
Send PM
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink]
1
MW + WW (if no member can hold two offices simultaneously - means repetition not allowed)

MW = 3/10 * 7/9 * 2! = 21/45

WW = 7/10 * 6/9 = 42/90

21/45 + 42/90 = 14/15
Prep Club for GRE Bot
[#permalink]
Moderators:
GRE Instructor
46 posts
GRE Forum Moderator
17 posts
Moderator
1090 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne