A biathlete shoots 5 times trying to hit a target. The probability to the the target is 0.8. Find the probability that he misses at least one of the 1st 2nd and 5th shot

Answer 1

I believe that what is needed here is the principle of inclusion exclusion.

A=miss on shot 1

B=miss on shot 2

C=miss on shot 3

P(A)=P(B)=P(C)=.2

P(A and B)=P(A and C)=P(B and C) = .04

P(A and B and C) = .008

P(at least one of A or B or C) =3*.2-3*.04+.008=.488

Look at it a different way.

Of the 32 possible sequences, 16 have a miss on the first shot.

Of the remaining 16, 8 have a miss on the 2nd shot

Of the remaining 8, 4 have a miss on the 5th shot.

Of the 4 remaining, there is one with 5 hits, 2 with 4 hits and 1 with 3 hits.

The probability of this remaining set of 4 is .8⁵ + 2*.8⁴*.2 + .8³*.2² = .512 and the probability of all the other possibilities is 1-.512=.488