Question

(10 pts) Alice and Bob are playing a dart game. Each person shoots 3 times. Assume all shots
are independent and Alice hits the target with probability 0.8 for each shot, while Bob hits the
target with probability 0.6. Calculate the probabilities of the following events.
(a) Alice hits the target no less than twice.
(b) Bob hits the target for the first time in the third shot.
(c) Alice hits the target more times than Bob.
(d) Alice hits the target exactly 2 more times than Bob.
(e) The total number of times Alice and Bob hit the target (out of 6 shots) is exactly 5.

Answer 1

a) P(Alice hits the target no less than twice) = P(Alice hits twice or thrice)

= 3x0.2x0.8² + 0.8³

= 0.384 + 0.512

= 0.896

b) P(Bob hits first time on third shot) = 0.4²x0.6

= 0.096

c) P(Alice hits the target more times than Bob) = P(Alice 1 and Bob 0) + P(Alice 2 and Bob 0 or 1) + P(Alice 3 and Bob 0 or 1 or 2)

= 3x0.8x0.2²x0.4³ + 3x0.8²x0.2x0.4³ + 3x0.8²x0.2x3x0.6x0.4² + 0.8³x0.4³ + 0.8³x3x0.6x0.4² + 0.8³x3x0.6²x0.4

= 0.006 + 0.025 + 0.111 + 0.033 + 0.147 + 0.221

= 0.543

d) P(Alice hits the target exactly 2 more times than Bob) = P(Alice 2 and Bob 0) + P(Alice 3 and Bob 1)

= 3x0.8²x0.2x0.4³ + 0.8³x3x0.6x0.4²

= 0.025 + 0.147

= 0.172

e) P(total number of times Alice and Bob hit the target is exactly 5) = P(Alice 3 and Bob 2) + P(Alice 2 and Bob 3)

= 0.8³x3x0.6²x0.4 + 3x0.8²x0.2x0.6³

= 0.221 + 0.083

= 0.304