Bernoulli distribution.
Exactly 2 times:
P(2)=C82⋅(41)2⋅(1−41)8−2≈0.311
3 times or more:
P(0)=C80⋅(41)0⋅(1−41)8−0≈0.1
P(1)=C81⋅(41)1⋅(1−41)8−1≈0.267
P(≥3)=1−P(0)−P(1)−P(2)≈1−0.1−0.267−0.311=0.322
Ali shots and Baba shots are independent events. Still Bernoulli distribution.
PAli(0)=C30⋅(41)0⋅(1−41)3−0≈0.422PAli(1)=C31⋅(41)1⋅(1−41)3−1≈0.422PAli(2)=C32⋅(41)2⋅(1−41)3−2≈0.141PAli(3)=C33⋅(41)3⋅(1−41)3−3≈0.016
PBaba(0)=C30⋅(31)0⋅(1−31)3−0≈0.296PBaba(1)=C31⋅(31)1⋅(1−31)3−1≈0.444PBaba(2)=C32⋅(31)2⋅(1−31)3−2≈0.222PBaba(3)=C33⋅(31)3⋅(1−31)3−3≈0.037
P=PAli(0)⋅PBaba(0)+PAli(1)⋅PBaba(1)+PAli(2)⋅PBaba(2)+PAli(3)⋅PBaba(3)≈≈0.422⋅0.296+0.422⋅0.444+0.141⋅0.222+0.016⋅0.037≈≈0.344
Ali takes 5 shots. So he misses 4 times and then hits the traget.
P=(1−41)⋅(1−41)⋅(1−41)⋅(1−41)⋅41≈0.079
Baba takes 4 shots. So when he shoots 4th time he hits the target 2nd time. First 3 shots can create any sequence of 1 hit and 2 missings. Bernoulli distribution again.
P=(C31⋅(31)1⋅(1−31)2)⋅31≈0.148
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