## Probability Worksheet Solution10

In this page probability worksheet solution10 we are going to see solution of some practice questions of the probability worksheet.

Question 6

A bag contains 50 bolts and 150 nuts. Half of the bolt and half of the nuts are rusted. If an item is chosen at random,find the probability that it is rusted or that it is a bolt.

Solution:

Sample space = 50 bolts + 150 nuts

= 200

Number of bolts rusted = 25

Number of nuts rusted = 75

Let A be the event of getting an item that it is a rusted item

A = 25 bolts + 75 nuts

n (A) = 100

P (A) = n (A)/n (S)

= 100/200

Let B be the event of getting an item that it is a bolt

n (B) = 50

P (B) = n (B)/n (S)

= 50/200

25 bolts are in both the events A and B

n (A ∩ B) = 25

P (A ∩ B) = n (A ∩ B)/n (S)

= 25/200

the probability that it is rusted or that it is a bolt

P (A U B) = P (A) + P (B) - P (A ∩ B)

= (100/200) + (50/200) - (25/200)

= (100 + 50 - 25)/200

= (150 - 25)/200

= 125/200

P (A U B) = 5/8

Question 7

Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the faces neither divisible by 3 nor by 4.

Solution:

when two dice are rolled,then the sample space (S) =
{(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)

(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)

(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)

(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)

(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)

(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}

n (S) = 36

To find the probability that the sum of the numbers on the faces neither divisible by 3 nor by 4 first we have to find the probability that the sum of numbers on faces are divisible by 3 or 4.

Let A be the event of getting a number divisible by 3

A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(4,5),(5,4),(3,3),(3,6),(6,3),(6,6)}

n (A) = 12

P (A) = n (A)/n (S)

= 12/36

Let B be the event of getting a number divisible by 4

B = {(1,3),(3,1),(2,2),(2,6),(3,5),(4,4),(5,3),(6,2),(6,6)}

n (B) = 9

P (B) = n (B)/n (S)

= 9/36

(A ∩ B) = {(6,6)}

n (A ∩ B) = 1

P (A ∩ B) = n (A ∩ B)/n (S)

= 1/36

P (A U B) = P (A) + P (B) - P (A ∩ B)

= (12/36) + (9/36) - (1/36)

= (12 + 9 - 1)/36

= 20/36

Probability that the sum of the numbers on the faces neither divisible by 3 nor by 4

P (A' ∩ B') = P(A U B)'

= 1 - P (A U B)

= 1 - (20/36)

= (36 - 20)/36

= 16/36              probability worksheet solution10

P (A' ∩ B') = 4/9

probability worksheet solution10 probability worksheet solution10