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.
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 )
A boy has an empty box. He flips a coin three times. Each time the coin...
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