Question

It is given that in a group of 3 students the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer 1

Hint: Here we use the concept of probability that probability of not occurring of an event is 1 minus the probability of occurring of an event. Use the given probability and find the required probability.
* Probability of an event is given by dividing the number of favorable outcomes by total number of outcomes. If we have an event A, the probability of which is given by \[P(A)\]; then probability of not that event i.e. A’ will be \[P(A') = 1 - P(A)\].

Complete step-by-step answer:
Here we are given a group of three students.
Let us denote the event of 2 students having same birthday as X
Then we have to find the value of\[P(X)\].
Probability of 2 students not having the same birthday\[ = 0.992\]
We can denote the event of 2 students not having the same birthday by X’
\[ \Rightarrow P(X') = 0.992\].............… (1)
Use the formula of probability, \[P(A') = 1 - P(A)\]
Substitute the value of \[X = A\]in the formula
\[ \Rightarrow P(X') = 1 - P(X)\]
Substitute the value of \[P(X') = 0.992\] from equation (1)
\[ \Rightarrow 0.992 = 1 - P(X)\]
Shift all constant values to one side of the equation.
\[ \Rightarrow P(X) = 1 - 0.992\]
Calculate the value in RHS of the equation.
\[ \Rightarrow P(X) = 0.008\]

\[\therefore \]Probability of 2 students having the same birthday is 0.008.

Note: Students might try to solve the probability using the general method by division of favorable outcomes by total outcomes. But this will be wrong as the number of students not sharing birthday i.e. the favorable outcome in the given case will come out to be in decimal form and we have a count of 1, 2 and 3 students. So we directly use the formula of probabilities sum here.