Question

A die is constructed in such a way that a 1 or 2 occurs twice as often
as a 5, which occurs three times as often as a 3,4, or 6. If the die is
tossed once, find the probability that
(a) the number is even;
(b) the number is a perfect square;
(c) the number is greater than 4.

Answer 1

Solution

Let the probability of getting 5 be y

i.e P(X=5) = y.

The die has the following probability distribution.

$\sum P(X=x) =1$

$$\therefore 6y+6y+y+y+3y+y=1$$ $$18y=1$$

$$y = {1 \over 18}$$

Substituting y with 1/18, The probability distribution is thus :

a). Probability the number is even.

$$=P(2 \cup 4 \cup 6)$$ $$= P(X=2) + P(X=4) + P(X=6)$$ $$= {1 \over 3} +{1 \over 18}+{1\over 18} = {4 \over 9}$$

Ans $4 \over 9$

b). Probability number is a perfect square.

$$P(X=1\space or \space X= 4) = P(X=1)+P(X=4)$$ $$={1 \over 3} + {1 \over 18} = {7 \over 18}$$

Ans: $7 \over 18$

c). Prob number is greater than 4

$$P(X>4)=P(X=5)+P(X=6)$$ $$= {1 \over 6} +{ 1 \over 18} = {2 \over 9}$$

Ans: $2 \over 9$