Each person in Room A is a student, and 1/6 of the students in Room A

Question

Each person in Room A is a student, and 1/6 of the students in Room A are seniors. Each person in Room B is a student, and 5/7 of the students in Room B are seniors. If 1 student is chosen at random from Room A and 1 student is chosen at random from Room B, what is the probability that exactly 1 of the students chosen is a senior?

(A) 5/42
(B) 37/84
(C) 9/14
(D) 16/21
(E) 37/42

Kudos for a correct solution.

(A) 5/42
(B) 37/84
(C) 9/14
(D) 16/21
(E) 37/42

Kudos for a correct  solution .

noTh1ng · Accepted Answer

We want the Probability of getting one senior and i.e. one junior. This is possible in two different cases: One senior from group A and no senior from group B or vice versa.

P(Senior A & No Senior B):
 \frac{1}{6} * \frac{2}{7} = \frac{1}{21}

P(No Senior A & Senior B):
 \frac{5}{7}*\frac{5}{6}=\frac{25}{42}

Add these two cases:
 \frac{1}{21}+\frac{25}{42} = \frac{9}{14}

Answer C

KS15 · Answer

Probability of choosing senior from Room A=1/6, Non senior=5/6
Probability of choosing senior from Room B=5/7, Non senior=2/7
Reqd probability= (1/6*2/7)+(5/7*5/6)=27/42=9/14
Answer C

kunal555 · Answer

Probability of choosing exactly 2 seniors from both the rooms is  \frac{1}{6}*\frac{5}{7}  =  \frac{5}{42}

Probability of not choosing any seniors from both the rooms is  \frac{5}{6}*\frac{2}{7}  =  \frac{10}{42} 
So,
Probability of choosing exactly 1 senior from each room will be

1- (  \frac{5}{42}  +  \frac{10}{42} )

= 1-  \frac{15}{42}

=  \frac{27}{42}

=  \frac{9}{14}

Answer:- C

anudeep133 · Answer

Solution:

Let A be an event of selecting a senior student from class A and B be an event of selecting a senior student from class B.
P(Exactly 1 senior Student) = P(A)P(Not B) + P(Not A)P(B) = (1/6)(2/7) + (5/6)(5/7) = 9/14

Option C

harishbiyani8888 · Answer

Probability of choosing senior from room A=1/6, Non senior=5/6

Probability of choosing senior from room B=5/7, Non senior=2/7

Required probability= (1/6*2/7)+(5/7*5/6)=27/42=9/14

Correct answer C

Bunuel · Answer

KAPLAN OFFICIAL SOLUTION:

In this problem we are asked to determine the probability of choosing exactly 1 senior. To be more specific, this means that we will need to select 1 senior and 1 non-senior.

There are two ways this can be done. We can select a senior from Room A and a non-senior from Room B or we can select a non-senior from Room A and a senior from Room B. In probability, ‘and’ means multiply while ‘or’ means add. In this case, we see 2 ‘ands’ indicating that we will multiply probabilities together and one ‘or’ telling us that we will add the resulting products.

Our first option is to get a senior from Room A, which has a probability of 1/6, and a non-senior from Room B, which has a probability of 2/7 (we found this number by taking the probability of selecting a senior in Room B and subtracting it from 1.) Because this is an ‘and’ situation, we multiply 1/6 * 2/7, which equals 2/42.

For our other option we must find the probability of selecting a non-senior from Room A, which has a probability of 5/6, and senior from Room B, which has a probability of 5/7. Again, we have an ‘and’ statement, so we multiply. 5/6 * 5/7 = 25/42.

Lastly, we must address the ‘or’ in our original scenario by adding the two probabilities we determined together. This gives us 2/42 + 25/42 = 27/42. This simplifies to 9/14, which is  choice (C) .

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Each person in Room A is a student, and 1/6 of the students in Room A

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