Probability Problems and Solutions

Definition:
probability is the number of favorable outcomes out of the number of possible outcomes.

Q-1) What is the probability of head side in the cricket toss ?
Solution:
Total side = (1 Head + 1 Tail) = 2
So, probability of head is = No. of Heads / Total Side
= 1/2
= 0.5
= 50% 
P(even) = 50%

Q-2) What is the probability of tail side in the cricket toss ?
Solution:
Total side =  (1 Head and 1 Tail) = 2  
So, probability of Tail is = No. of Tails / Total Side
= 1/2
= 0.5
= 50% 
P(tail) = 50%

Q-3) What is the probability of even number when you roll the die ?
Solution:
Total number in a die =  1,2,3,4,5,6 = 6 numbers
Total even numbers  = 2,4,6 = 3 numbers
= 3/6
= 1/2
= 0.5
=50%
P(evenNumber) = 50%

Q-4) What is the probability of odd number when you roll the die ?
Solution:
Total number in a die =  1,2,3,4,5,6 = 6 numbers
Total even numbers  = 1,3,4 = 3 numbers
= 3/6
= 1/2
= 0.5
=50%
P(oddNumber) = 50%

Q-5) What is the probability of prime number when you roll the die ?
Solution:
(Note: A number divisible by itself only is called prime number)
Total number in a die =  1,2,3,4,5,6 = 6 numbers
Total prime numbers  = 1,3,5 = 3 numbers

= 3/6
= 1/2
= 0.5
=50%
P(prime) = 50%

Q-6) Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a divisible by 3 ?
Solution:
Total Numbers = 20
Numbers divisible by 3  = 6,9,12,15,18 = 5 Numbers
P(number divisible by 3) =  5/20 
= 1/4 
P(number divisible by 3) = 25%

Q-7) Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a divisible by 3 and 6?
Solution:
Total Numbers = 20
Numbers divisible by 3  = 6,9,12,15,18 = 5 Numbers
Numbers divisible by 5  = 5,10,15,20 = 4 Numbers
Total Unique number divisible by 3 and 5
= 6,9,10,12,15,18,20

P(number divisible by 3 and 5) 
=  7/20 
= 0.35 
P(number divisible by 3 and 5)  = 35%

Q-8) A bag contains 4 red, 5 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Solution:
Total number of balls = (4 + 5 + 2) = 11.
Let S be the space.
Then, n(S)= Number of ways of drawing 2 balls out of 11
=11C2 
=(11 x10)/(2 x 1)
=110/2
=55

Let E = Drawing 2 balls, none of which is blue.
n(E)= Number of ways of drawing 2 balls out of (4 red + 5 green)9 balls
=9C2
=(9 x 8)/(2 x 1)
=36
P(E)=n(E)/n(S)
=36/55
probability that none of the balls drawn is blue = 36/55

Q-9) In a bucket, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
Solution:
Total number of balls 
= (8 + 7 + 6) 
= 21
n(S) = 21

Let E = An event the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7

P(E) = n(E)/n(S)
 = 7/21
P(E) = 1/3
probability that a ball drawn is neither red nor green = 1/3

Q-10) What is the probability of getting a sum 9 from two throws of a dice?
Solution:
Two throw of dice 
n(S) = (6 x 6) = 36

Let E = event of getting a sum 9 in two throws 
={(3, 6), (4, 5), (5, 4), (6, 3)}

P(E) = n(E)/n(S)
= 4/36
P(E)= 1/9