Question

Two dice are rolled once. Find the probability of getting such numbers on two dice, whose product is a perfect square.

OR

A game consists of tossing a coin 3 times and noting its outcomes each time. Hanif wins if he gets three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer 1

The sample space of the given experiment is:
$S=\left\{\begin{array}{cccccc}(1,1),& (1,2),& (1,3),& (1,4),& (1,5),& (1,6)\\ (2,1),& (2,2),& (2,3),& (2,4),& (2,5),& (2,6)\\ (3,1),& (3,2),& (3,3),& (3,4),& (3,5),& (3,6)\\ (4,1),& (4,2),& (4,3),& (4,4),& (4,5),& (4,6)\\ (5,1),& (5,2),& (5,3),& (5,4),& (5,5),& (5,6)\\ (6,1),& (6,2),& (6,3),& (6,4),& (6,5),& (6,6)\end{array}\right\}$
$\therefore$ n (S)= 36
Let E be the event that the product of numbers obtained on the upper face is a perfect square
$\therefore$ E = {(1,1), (1,4), (2,2), (3,3), (4,1), (4,4), (5,5), (6,6)}
$\therefore$ n (E) = 8
$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{8}{36}=\frac{2}{9}$
OR
The set of possible outcomes of the given experiment are:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Let E be the event of getting three heads or three tails.
$\therefore$ E = {HHH, TTT}
$\therefore$ Probability of winning = P (E)
$=\frac{n\left(E\right)}{n\left(S\right)}=\frac{2}{8}=\frac{1}{4}$
$\therefore$ Probability of losing = P(E')
= 1 - P (E)
$=1-\frac{1}{4}=\frac{3}{4}$
Therefore, the probability that Hanif will lose the game is $\frac{3}{4}$