Question 37632: 1o balls numbered 1 to 10 are in a jar. Jack reaches into jar & randomly removes one of the ball. Then jill reaches into the jar & randomlly removes a different ball..What is the probability that the sum of the 2 numbers on the balls removed is even?
Answer by mszlmb(115)   (Show Source):
You can put this solution on YOUR website! ok odd+odd=even even+even=even odd+even=odd
so we're looking at the probability that both picked either odd numbers or that both chose even numbers
too hard, let's just take the total probability (1) and subtract from it the probability that 1 picked odd and 1 even.
p(1st guy picking even)or p(1st guy picking odd)=.5
cuz there's 5/10 even numbers (2,4,6,8,10) and 5/10 odd (1,3,5,7,9)
p(2nd guy picking odd)or p(2nt guy picking even)=.5
same reason
p(2nd guy picking the opposite of 1st guy)=5/10*5/9=5/18=.25252525....
5/10 cuz 5 of 10 of choices r either even or odd and
1/9 cuz of those 5/10, 5/9 of the balls remaining were the opposite.
we times them cuz we have 5/10ths a chance @ first and of those 5/10ths n array of 5/9ths per 5/10ths.
NOTE had Jack (the 1st dude) put the ball back, p(2nd guy picking opposite of 1st guy) would be 5/10*5/10, but because he takes the ball out, Jill now has a 5/9ths chance to pick a ball opposite his.
if uv bin keeping up, the answer .25252525... is the probability that the sum of the numbers is not even.
1-.25252525...=.7474747474.. or .75. 75% chance, .75 probability, 75/100.
chek and check^2 to c if im right, and plz email me if im wrong.
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