A bag contains 100 balls numbered from 1 to 100. If 3 balls are selected randomly from the bag with replacement, what is the probability that sum of the numbers on 3 balls is odd ?
a)3/8
b)1/2
c)5/8
d)1/4
e)3/4
Approach 1:
The sum will be odd if all 3 numbers selected are odd or if exactly 1 number selected is odd.
Let O = odd and E = even.
Since there are an equal number of odd numbers and even numbers in the box, P(O) = 1/2 and P(E) = 1/2.
Probability all 3 are odd:
P(OOO) = 1/2 * 1/2 * 1/2 = 1/8
Probability exactly 1 is odd:
P(OEE) = 1/2 * 1/2 * 1/2 = 1/8
Since EOE and EEO are also good outcomes, we multiply this result by 3:
3 * 1/8 = 3/8
Since we'll get a good outcome if all 3 numbers selected are odd OR if exactly 1 number selected is odd, we add the fractions:
1/8 + 3/8 = 4/8 = 1/2.
The correct answer is
B.
Approach 2:
Since there are 50 odd numbers and 50 even numbers:
P(odd number) = 1/2.
P(even number) = 1/2.
Ways to get an ODD sum:
3 odd numbers
1 odd number, 2 even numbers
Ways to get an EVEN sum:
3 even numbers
1 even number, 2 odd numbers
Since P(odd) = P(even) = 1/2:
P(3 odds) = P(3 evens).
P(1 odd, 2 evens) = P(1 even, 2 odds).
Adding the two equations, we get;
P(3 odds) + P(1 odd, 2 evens) = P(3 evens) + P(1 even, 2 odds)
P(odd sum) = P(even sum).
Since the two probabilities are equal, and the sum must be odd or even:
P(odd sum) = 1/2.
P(even sum) = 1/2.
The correct answer is
B.
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