11. A couple have 3 children. None of the children were born in a leap year. The probability that all three children share the same birthday is: a. 3/365 b. (1/365)2 c. (1/365)3 d. (1/365) x (1/364) x (1/363)

11. A couple have 3 children. None of the children were born in a leap year. The
probability that all three children share the same birthday is:
a. 3/365
b. (1/365)2
c. (1/365)3
d. (1/365) x (1/364) x (1/363)

In probability theory, a classic question asks for the probability that three children have the same birthday in a non‑leap year, assuming each day of the year is equally likely. This concept is closely related to the famous birthday problem and helps students understand how to apply the multiplication rule for independent events in discrete probability.

Correct Answer

The correct answer is option b. (1/365)2. The probability that all three children share the same birthday (same day and month) in a non-leap year is
1 / 365^2.

Step-by-Step Solution

There are 365 possible birthdays in a non‑leap year (ignoring the year of birth and assuming all days are equally likely).

  • The birthday of the first child can be any day of the year, so no restriction or probability factor is needed for the first child.
  • The second child must have exactly the same birthday as the first child. The probability of this is 1/365.
  • The third child must also have exactly the same birthday as the first child. The probability of this is again 1/365.

Using the multiplication rule for independent events, the probability that both the second and third child match the first child is:

P(all 3 share same birthday)
= (1/365) × (1/365)
= 1 / 365²

Therefore, the required probability is (1/365)2.

Explanation of Each Option

Option Expression Meaning Correct? Reason
a 3/365 As if “any one of three” matches some fixed day. No This is linear in 1/365 and does not use the multiplication rule for joint probability of all three matching.
b (1/365)2 Second matches first, third matches first, as independent events. Yes Correct use of independence; probability that both the second and third child match the first child is (1/365) × (1/365) = 1/365².
c (1/365)3 All three matching some pre‑chosen specific calendar date. No This would be correct only if the question specified a particular date in advance (for example, all three are born on 1st January).
d (1/365) × (1/364) × (1/363) Pattern used for “all different birthdays”. No This structure corresponds to the probability that no two share a birthday, not that all three have identical birthdays.
 

 

Leave a Reply

Your email address will not be published. Required fields are marked *

Latest Courses