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27 Apr 2025, 05:59
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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, 14:23
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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!
