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Five balls are numbered with the integers 1 through 5 and placed in a jar. Three are drawn without replacement. What is the probability that the sum of the three integers on the balls is odd? Express your answer as a common fraction.

Nov 2, 2018

#1
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The balls can sum to odd if they are

(o, o, o), (o, e, e)

There is only 1 way to choose (o,o,o), i.e. (1,3,5)

There are 3 ways to choose (o,e,e), both the two even numbers, 2 and 4, and one of the odd numbers.

Thus there are 4 total ways 3 balls can sum to an odd number.

There are $$\dbinom{5}{3} = 10$$ total ways to select 3 balls from the 5.  Thus

$$P[\text{choose 3 balls that sum to an odd number}]=\dfrac{4}{10}=\dfrac 2 5$$

.
Nov 3, 2018

#1
+5078
+1

The balls can sum to odd if they are

(o, o, o), (o, e, e)

There is only 1 way to choose (o,o,o), i.e. (1,3,5)

There are 3 ways to choose (o,e,e), both the two even numbers, 2 and 4, and one of the odd numbers.

Thus there are 4 total ways 3 balls can sum to an odd number.

There are $$\dbinom{5}{3} = 10$$ total ways to select 3 balls from the 5.  Thus

$$P[\text{choose 3 balls that sum to an odd number}]=\dfrac{4}{10}=\dfrac 2 5$$

Rom Nov 3, 2018