Question

A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?

Answer 1

Hint- Number of favourable outcomes divided by total number of outcomes gives probability.Sample space is defined as set of all possible outcomes or results of that experiment.For throwing a die twice, the sample space considered as:
Sample space = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
                                   (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
                                   (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
                                   (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
                                    (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
                                   (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Complete step-by-step answer:
According to question,
We need to see favorable conditions when 5 will not come up either time, we get
(1,1), (1,2), (1,3), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,6), (3,1), (3,2), (3,3), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,6), (6,1), (6,2), (6,3), (6,4), (6,6)
We know definition of probability i.e. Number of favourable outcomes divided by the total number of outcomes gives probability.
Here, Number of favourable outcomes =25 and Total number of outcomes =36
(i) P (5 will not come up either time) = Number of times 5 does not come divided by total number of outcomes
P (5 will not come up either time) = $\dfrac{{25}}{{36}}$

Again, we need to see the favourable conditions when 5 will come up at least once, we get
(1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,5)
 Here, Number of favourable outcomes =11 and Total number of outcomes =36
(ii) P (5 will come up at least once) = Number of times 5 comes at least once divided by total number of outcomes
P (5 will come up at least once) = $\dfrac{{11}}{{36}}$
Note- Throwing a die twice and throwing two dice simultaneously are treated as the same experiment. Hence the sample space remains the same as mentioned above.