Question
Question: Reddy has two pairs of black shoes and three pairs of brown shoes. He also has three pairs of red so...
Reddy has two pairs of black shoes and three pairs of brown shoes. He also has three pairs of red socks, four pairs of brown socks and six pairs of black socks. If Reddy chooses a pair of shoes at random and a pair of socks at random . Using a tree diagram, find the probability that he chooses shoes and socks of the same colour.
(A) 134
(B) 6524
(C) 135
(D) 136
Solution
If the probability of occurring an event is ′a′ and occurring an independent event is ′b′ , then the probability of both the event occurring is ab. We can use this formula separately to find the probability of getting black shoes-black socks and brown shoes-brown socks.
Complete step-by-step answer:
Given, Reddy has 2 pairs of black shoes and 3 pairs of brown shoes.
Therefore, total no. pairs of shoes =5
The probability of an event is given by,P(A)=n(S)n(A)
Where P(A) is the probability of an event A, n(A) is the no. of favorable outcomes and n(S) is the total no. of outcomes.
Now, probability of getting black shoes =52
& Probability of getting brown shoes =53
Also given that Reddy has 3 pairs of red socks, 4pairs of brown socks and 6 pairs of black socks.
Therefore, total no. of socks =13
Now, probability of getting red socks =133
Probability of getting brown socks =134
& Probability of getting black socks =136
Now, probability of getting shoes and socks of the same colour = (Probability of getting black shoes ×Probability of getting black socks) + (Probability of getting black shoes ×Probability of getting black socks)
∴P=P(All black)+P(All brown)
⇒P = 52×136+53×134
⇒P = 6512+6512
⇒P = 6524
Therefore, the probability that Reddy chooses shoes and socks of the same colour is 6524.
Hence, option (B) is the correct answer.
Note: In the given problem, there is no pair of red shoes. So, the probability of getting red shoes will be zero and hence the probability of getting shoes and socks of the same red colour will also be zero. Therefore, we eliminate the P(All red), while finding the final solution.