Question

A die with number 1 to 6 is biased such that P(2) = `3/10` and probability of other numbers is equal. Find the mean of the number of times number 2 appears on the dice, if the dice is thrown twice.

Answer 1

Total outcomes, S = {1, 2, 3, 4, 5, 6}

Given: P(2) = `3/10`

P(1) = P(3) = P(4) = P(5) = P(6) = K (let)

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1

`K + 3/10 + K + K + K + K = 1`

`5K + 3/10 = 1`

50 K + 3 = 10

50 K = 10 − 3

50 K = 7

K = `7/50`

q(not getting 2)

= 1 − P(2)

= `1 - 3/10`

= `(10 - 3)/10`

= `7/10`

Let X→ number of time 2 appears 

X = 0, 1, 2

P(X = 0) = P(both not 2)

= `7/10 xx 7/10`

= `49/100`

P(X = 1) = 2P(one 2 and other not 2)

= `(2 xx 3)/10 xx 7/10`

= `42/100`

P(X = 2) = P(both 2)

= `3/10 xx 3/10`

= `9/100`

For Mean:

X	P	XP
0	`49/100`	0
1	`42/100`	`42/100`
2	`9/100`	`18/100`

Mean = E(x) 

`sumx_i P_i = 0 + 42/100 + 18/100`

= `(42 + 18)/100`

= `60/100`

= 0.6