Events & Promotions
|
|
GMAT 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1410:00 AM PDT
-11:59 PM PDT
GMAT Preparation Online - A limited time sale use code: EXPERTSGLOBAL10- Jul 15
08:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations! 
Jul 1609:30 AM EDT
-11:30 AM EDT
Is INSEAD your dream b-school and you're looking for solid prep and application strategies to get there? Check out Barkavi’s end-to-end journey to know how she did! Get insights into the INSEAD admissions process, know what Adcoms look at, and more!
Jul 1611:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
Jul 1911:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
27 Apr 2025, 05:59
Kudos
Bookmarks
Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls.
What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)
Total ways = 2 boys 2 girls + 3 girls 1 boy + 4 girls = 4!/2!2!+ 4!/3! + 1 = 6+4+1=11
Favorable ways = 2 boys 2 girls = 4!/2! = 6
The required probability = 6/11
IMO E
What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)
Total ways = 2 boys 2 girls + 3 girls 1 boy + 4 girls = 4!/2!2!+ 4!/3! + 1 = 6+4+1=11
Favorable ways = 2 boys 2 girls = 4!/2! = 6
The required probability = 6/11
IMO E
08 Jun 2025, 11:51
Kudos
Bookmarks
isnt p(a given b)=p(A intersection b)/p(b)
but i guess since event a comes under event b therer intersection is event a thus p(A)?
The following is the complete explanation for this question:
You need to find Probability of 2 boys and 2 girls given there are at least 2 girls. This is a conditional probability. You have already been given that there are at least 2 girls. You need to find the probability of 2 boys and 2 girls given this condition.
P(A given B) = P(A)/P(B)
P(2 boys, 2 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8
You multiply by 4!/(2!*2!) because out of the four children, any 2 could be boys and the other two would be girls. So you have to account for all arrangements: BGBG, BBGG, BGGB etc
P(atleast 2 girls) = P(2 boys, 2 girls) + P(1 boy, 3 girls) + P(4 girls) = 1 - P(4 boys) - P(3 boys, 1 girl)
You can solve P(atleast 2 girls) in any way you like.
P(1 boy, 3 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4
P(4 girls) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16
P(atleast 2 girls) = 3/8 + 1/4 + 1/16 = 11/16
P(2 boys, 2 girls given atleast 2 girls) = P(2 boys, 2 girls) / P(atleast 2 girls) = (3/8)/(11/16) = 6/11
In the above given solutions, all the (1/2)s have been ignored because the probability of a boy is 1/2 and of a girl is 1/2 too. Hence in every case you will get (1/2)*(1/2)*(1/2)*(1/2) which can be ignored. We just need to focus on arrangements in each case. This is similar to the famous coin flipping questions (the probability of 3 Heads and a tail etc)
but i guess since event a comes under event b therer intersection is event a thus p(A)?
KarishmaB
raghupara
Hi people,
Suddenly, something comes to my mind so awfully, but I couldn't disprove why I am wrong. Could someone help me in this.
I suddenly see it as : P(2 girls, 2 boys) = 1/2*1/2*1/2*1/2 = 1/16.(since each has a prob of 1/2).
Pls disprove me.... I don see where I am wrong, but I know surely I am.
Thanks
Suddenly, something comes to my mind so awfully, but I couldn't disprove why I am wrong. Could someone help me in this.
I suddenly see it as : P(2 girls, 2 boys) = 1/2*1/2*1/2*1/2 = 1/16.(since each has a prob of 1/2).
Pls disprove me.... I don see where I am wrong, but I know surely I am.
Thanks
The following is the complete explanation for this question:
You need to find Probability of 2 boys and 2 girls given there are at least 2 girls. This is a conditional probability. You have already been given that there are at least 2 girls. You need to find the probability of 2 boys and 2 girls given this condition.
P(A given B) = P(A)/P(B)
P(2 boys, 2 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8
You multiply by 4!/(2!*2!) because out of the four children, any 2 could be boys and the other two would be girls. So you have to account for all arrangements: BGBG, BBGG, BGGB etc
P(atleast 2 girls) = P(2 boys, 2 girls) + P(1 boy, 3 girls) + P(4 girls) = 1 - P(4 boys) - P(3 boys, 1 girl)
You can solve P(atleast 2 girls) in any way you like.
P(1 boy, 3 girls) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4
P(4 girls) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16
P(atleast 2 girls) = 3/8 + 1/4 + 1/16 = 11/16
P(2 boys, 2 girls given atleast 2 girls) = P(2 boys, 2 girls) / P(atleast 2 girls) = (3/8)/(11/16) = 6/11
In the above given solutions, all the (1/2)s have been ignored because the probability of a boy is 1/2 and of a girl is 1/2 too. Hence in every case you will get (1/2)*(1/2)*(1/2)*(1/2) which can be ignored. We just need to focus on arrangements in each case. This is similar to the famous coin flipping questions (the probability of 3 Heads and a tail etc)
08 Jun 2025, 14:23
Kudos
Bookmarks
Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)
(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11
Hello, people! Interesting question a person may want to be a bit careful with. There's a combination vid on the gmatknight blog people can also check out that is sort of similar but this is nice, too. For this question, one important thing to note is that we already know she has at least 2 boys, so we're not dealing with a probability out of every possible combination of 4 children. For example, Boy, Boy, Boy, Boy is not something we need to consider.
So let's get stuck in. Let's figure out the total number of potential combinations in which she has at least 2 girls.
Step 1: Let's start with the number of variations of her having EXACTLY 2 girls (e.g. G, G, B, B):
Remember, we can shuffle this anyway we want so: 4! / 2! X 2! = 6
Step 2: Let’s imagine all the possibilities she has EXACTLY 3 children (e.g. G, G, G, B):
Again, we can shuffle, so: 4! / 3! = 4
Step 3. And finally exactly 4 girls (G, G, G, G):
Clearly, we can only have 1 combination like this.
So, all in all, the total number of variations in which she has at least 2 girls is (6 + 4 + 1) = 11
Now, all we need to know is the total number of combinations of her having exactly 2 boys. The great thing is we don't need to solve for this since basically this is the same as the number of combinations of her having exactly 2 girls, which we already solved for in Step 1.
(E) 6/11 is therefore your answer.
All the best!
(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11
Hello, people! Interesting question a person may want to be a bit careful with. There's a combination vid on the gmatknight blog people can also check out that is sort of similar but this is nice, too. For this question, one important thing to note is that we already know she has at least 2 boys, so we're not dealing with a probability out of every possible combination of 4 children. For example, Boy, Boy, Boy, Boy is not something we need to consider.
So let's get stuck in. Let's figure out the total number of potential combinations in which she has at least 2 girls.
Step 1: Let's start with the number of variations of her having EXACTLY 2 girls (e.g. G, G, B, B):
Remember, we can shuffle this anyway we want so: 4! / 2! X 2! = 6
Step 2: Let’s imagine all the possibilities she has EXACTLY 3 children (e.g. G, G, G, B):
Again, we can shuffle, so: 4! / 3! = 4
Step 3. And finally exactly 4 girls (G, G, G, G):
Clearly, we can only have 1 combination like this.
So, all in all, the total number of variations in which she has at least 2 girls is (6 + 4 + 1) = 11
Now, all we need to know is the total number of combinations of her having exactly 2 boys. The great thing is we don't need to solve for this since basically this is the same as the number of combinations of her having exactly 2 girls, which we already solved for in Step 1.
(E) 6/11 is therefore your answer.
All the best!
