RGS Practice Question
Two dice are thrown. One is biased so that the probability of throwing a 6 is 1/16 and the probabilities of throwing the other five numbers are all equal. The other die is unbiased. Calculate, as a fraction in its lowest term, the probability that
(i) both numbers thrown are not 6,
(ii) at least one of the two numbers thrown is a 6,
(iii) exactly one of the numbers thrown is a 4.
*************************
Answer:
(i) P(both numbers thrown are not 6)
= P(1st dice not 6 and 2nd dice not 6)
= 15/16 * 5/6
= 25/32
(ii) P(at least one of the two numbers thrown is a 6)
= 1 - P(both numbers thrown are not 6)
= 1 - 25/32
= 7/32
(iii) For the biased dice, the probability of getting 1, 2, 3, 4 or 5 is (1 - 1/16)/5 = 3/16
P(exactly one of the numbers thrown is a 4)
= P(1st dice is 4, and 2nd dice is not 4) + P(1st dice is not 4, and 2nd dice is 4)
= 3/16 * 5/6 + 13/16 * 1/6
= 5/32 + 13/96
= 7/24
