Question

A die is thrown twice. Find the probability of
(A) 5 coming up neither time
(B) 5 coming up at most once.

Answer 1

Hint : We first try to find the unconditional and conditional number of outcomes for the events of throwing the dice twice. We then use the probability theorem to find their individual probabilities.

Complete step by step solution:
A die is thrown twice. We have to find the probability of 5 coming up either time.
We take the whole event of throwing the dice twice as the unconditional event of S.
In each throw the number of outcomes for the dice is 6. The two throws are independent.
Therefore, the total number of outcomes is $6\times 6=36$ .
We take $n\left( S \right)=36$ .
Now we take the event of 5 coming up neither time as A.
So, the number of outcomes for each throw is 5. The total number of outcomes is $5\times 5=25$ .
So, $n\left( A \right)=25$ . The probability is
$p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{25}{36}$.
Now we take the event of 5 coming up at most once B.
So, the number of outcomes for one throw is fixed being 5. The total number of outcomes is $5\times 1\times 2+5\times 5=35$ . There are two choices for 5 to appear.
So, $n\left( B \right)=35$ . The probability is $p\left( B \right)=\dfrac{n\left( B \right)}{n\left( S \right)}=\dfrac{35}{36}$.
The respective probabilities are $\dfrac{25}{36},\dfrac{35}{36}$.
So, the correct answer is “$\dfrac{25}{36},\dfrac{35}{36}$.”.

Note : We could have broken the event of 5 coming up at most once in two parts where we use the complement event of 5 coming up only once as the opposite of 5 coming up twice.
So, $n\left( B \right)=36-1=35$ .