Asked by Charlottte
There are 9 numbers (1-9) placed in a bag, if two numbers are drawn without replacement in succession, find the probability that the sum of the numbers is odd?
Answers
Answered by
Reiny
For the sum to be odd, one choice must be even , the other odd
(the sum of two odds is even, the sum of 2 evens is even)
We have 5 odds and 4 evens in the bag
so 2 cases:
prob(E, then O) = (4/9)(5/8) = 20/72 = 5/18
prob(O, then E) = (5/9)(4/8) = 5/18
prob(your event) = 5/18+5/18 = 10/18 = 5/9
check:
two other cases:
Prob(E,E) = (4/9)(3/8) = 1/6
prob(O,O) = 5/9)(4/8) = 5/18
sum: 5/9 + 5/18 + 1/6 = 1 , YEAHH
(the sum of two odds is even, the sum of 2 evens is even)
We have 5 odds and 4 evens in the bag
so 2 cases:
prob(E, then O) = (4/9)(5/8) = 20/72 = 5/18
prob(O, then E) = (5/9)(4/8) = 5/18
prob(your event) = 5/18+5/18 = 10/18 = 5/9
check:
two other cases:
Prob(E,E) = (4/9)(3/8) = 1/6
prob(O,O) = 5/9)(4/8) = 5/18
sum: 5/9 + 5/18 + 1/6 = 1 , YEAHH
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