Question

A boy has an empty box. He flips a coin three times. Each time the coin lands heads, he puts a red marble in the box. Each time the coin lands tails, he puts a green marble in the box. He then reaches in the box and pulls out one of the three marbles at random.

• What is the chance that the boy puts exactly two green marbles in the box?
• What is the chance the boy puts at least one green marble in the box?
• What is the chance the boy pulls out a green marble from the box?
• Given that the coin lands heads exactly once, what is the chance that the boy pulls out a green marble from the box?
• Given that the boy pulls out a green marble from the box, what is the chance that the coin landed heads exactly once?
• Given that the boy pulls out a green marble from the box, what is the chance that the coin landed heads at least once?
• Given that the boy pulls out a green marble from the box, what is the chance that the coin landed tails at least once?

Answer 1

A boy tossed coin three times.

The possible way to fill the box

Coins outcome	Put the marble in Box	Number of red marble in a box (X)	Number of green marble in box (Y)
HHH	RRR	3	0
HHT	RRG	2	1
HTH	RGR	2	1
HTT	RGG	1	2
THH	GRR	2	1
THT	GRG	1	2
TTH	GGR	1	2
TTT	GGG	0	3

i) P ( Boy puts exactly two green marble in the box) = P(Y=2) = 3/8 = 0.375.

ii) P ( Boy puts at least one green marble in the box) = 1 - P ( Boy didn't puts any green marble)

= 1- P( Y=0) = 1- 1/8 = 0.875.

iii) A : P ( Boy pulls out a green marble from the box)

= 1/8 * ( 3 * 1/3 + 3 * 2/3 + 1 * 1)

= 4/ 8

= 0.5000

iv) B : Coin land heads exactly one

P ( B ) = 3/8

Exactly one head	No. of green marble in a box	P ( Drawing a green marble)
HTT	2	2/3
THT	2	2/3
TTH	2	2/3

Required Probability = P ( A / B )