Question

A box contains 10 balls. Five balls are numbered 0 and the other five balls are numbered 1, 2, 3, 4, 5, respectively. a Three balls are randomly chosen with replacement. What is the probability that the sum of the numbers on the chosen balls is 10

Answer 1

33/1000

Explanation:

Count the case to get 10

5,5,0= 16

4,4,2=3

5,2,3=6

5,4,1=6

4,3,3=3

15+6+6+3+3= 33 then 33/1000 will be our answer

Answer 2

1/100

Explanation:

The number of total possible outcomes is 10(we have ten balls)

let, P(O) be the probability of choosing a ball that is numbered O

P(1) be the probability of choosing a ball that is numbered 1

P(2) be the probability of choosing a ball that is numbered 2

P(3) be the probability of choosing a ball that is numbered 3

P(4) be the probability of choosing a ball that is numbered 4

P(5) be the probability of choosing a ball that is numbered 5

since probability = number of favorable outcome / number of total possible outcome

The favorable outcome for balls that are number 0 is 5, making P(O)= 5/10

The favorable outcome for balls that are number 1 is 1, making P(1)= 1/10

The favorable outcome for balls that are number 2 is 1, making P(2)= 1/10

The favorable outcome for balls that are number 3 is 1, making P(3)= 1/10

The favorable outcome for balls that are number 4 is 1, making P(4)= 1/10

The favorable outcome for balls that are number 5 is 1, making P(5)= 1/10

Three balls are randomly chosen with replacement, to find the probability of choosing three balls whose sum will make 10, we have to consider the possible combination of the sum of three numbers that could result to 10

we have :

0+5+5(since the balls were chosen with replacement ,we can pick two balls numbered 5 and a ball numbered 0)

1+4+5

2+3+5

2+4+4

3+3+4

Hence the probability will be given as ;

[p(0)×P(5)×P(5)] + [P(1)×P(4)×P(5)] + [P(2)×P(3)×P(5)] + [P(2)×P(4)×P(4)] + [P(3)×P(3)×P(4)] =

(1/2×1/10×1/10) + (1/10×1/10×1/10) + (1/10×1/10×1/10) + (1/10×1/10×1/10) + (1/10×1/10×1/10) = 1/100