Question

Couldn't find an easy way to do this question. Any advice is appreciated.

Suppose a,b,c,d,e are selected randomly from the set {1,2,3,4,5} and they can repeat. Find the probability that a*b*c*d + e is odd.

Thanks for the help!

Answer 1

We need to find P[ a*b*c*d + e is odd ]

there are two possibilities

a*b*c*d is odd and e is even or a*b*c*d is even and e is odd

P[ a*b*c*d + e is odd ] = P[ a*b*c*d is odd and e is even ] + P[ a*b*c*d is even and e is odd ]

P[ a*b*c*d + e is odd ] = P[ a*b*c*d is odd ]*P[ e is even ] + P[ a*b*c*d is even ]*P[ e is odd ]

a*b*c*d is even if even the one member of a*b*c*d is even else odd

P[ selecting even member from the set ] = 2/5

P[ selecting odd member from the set ] = 3/5

P[ e is even ] = 2/5

P[ e is odd ] = 3/5

P[ a*b*c*d is odd ] = (3/5)^4 = 81/625 ( all members are odd )

P[ a*b*c*d is even ] = 1 - P[ a*b*c*d is odd ] = 1 - 81/625 = 544/625

P[ a*b*c*d + e is odd ] =(81/625)*(2/5) + (544/625)*(3/5)

P[ a*b*c*d + e is odd ] = 162/3125 + 1632/3125

P[ a*b*c*d + e is odd ] = 1794/3125

P[ a*b*c*d + e is odd ] = 0.5741