Dear Reader,
Below are four different problems based on probability.
Question 1
In a class, there are 12 boys and 16 girls. One of them is called out by an enroll number, what is the probability that the one called is a girl?
a) 1/4 b) 2/5 c) 5/12 d) 4/7
Answer : d) 4/7.
Solution :
Let S be the sample space.
Total number of students in the class = 12 boys + 16 girls = 28.
Then, n(S) = 28.
Let E be the event of calling one of them by enroll number.
Given that, number of girls = 16.
Then, n(E) = 16.
The probability that the one called is a girl = n(S)/n(E) = 16/28 = 4/7.
Question 2
A box contains 3 white, 4 red and 7 blue erasers. If five erasers are taken at random then the probability that all the five are blue color is:
a) 13/211 b) 2/126 c) 3/286 d) 12/121
Answer : c) 3/286
Solution :
Total number of erasers in the box = 3 + 4 + 7 = 14.
Let S be the sample space.
Then, n(S) = number of ways of taking 5 out of 14.
Therefore, n(S) = 14C5 = (14 x 13 x 12 x 11 x 10)/(2 x 3 x 4 x 5) = 14 x 13 x 11 = 2002
Let E be the event of getting all the 5 blue erasers.
Therefore, n(E) = 7C5 = (7 x 6 x 5 x 4 x 3)/(2 x 3 x 4 x 5) = 21.
Now, the required probability = n(E)/n(S) = 21/2002 = 3/286.
Question 3
A box contains 6 bottles of variety1 drink, 3 bottles of variety2 drink and 4 bottles of variety3 drink. Three bottles of them are drawn at random, what is the
probability that the three are not of the same variety?
a) 833/858 b) 752/833 c) 632/713 d) none of these
Answer : a) 833/858
Solution :
Total number of drink bottles = 6 + 3 + 4 = 13.
Let S be the sample space.
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = 13C3 = (13 x 12 x 11)/(1 x 2 x 3) = 66 x 13 = 858.
Let E be the event of taking 3 bottles of the same variety.
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = 6C3 + 3C3 + 4C3
= 6 x 5 x 4 / 1 x 2 x 3 + 1 + 4 x 3 x 2 / 1 x 2 x 3
= 20 + 1 + 4 = 25.
The probability of taking 3 bottles of the same variety = n(E)/n(S) = 25/858.
Then, the probability of taking 3 bottles are not of the same variety = 1 - 25/858 = 833/858.
Question 4
There are 12 boys and 8 girls in a tution centre. If three of them scored first mark, then what is the probability that one of the three is a girl and the other two are boys?
a) 14/75 b) 22/55 c) 44/95 d) none of these
Answer : c) 44/95
Solution :
Total number of students = 20.
Let S be the sample space.
Then, n(S) = number of ways of three scored first mark
n(S) = 20C3 = 20 x 19 x 18 / 2 x 3 = 20 x 19 x 3
Let, E be the event of 1 girl and 2 boys.
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12.
n(E) = 8C1 x 12C2 = 8 x 12 x 11 / 1 x 2 = 8 x 6 x 11.
Now, the required probability = n(E)/n(S) = (8 x 6 x 11)/(20 x 19 x 3) = 44/95
Below are four different problems based on probability.
Question 1
In a class, there are 12 boys and 16 girls. One of them is called out by an enroll number, what is the probability that the one called is a girl?
a) 1/4 b) 2/5 c) 5/12 d) 4/7
Answer : d) 4/7.
Solution :
Let S be the sample space.
Total number of students in the class = 12 boys + 16 girls = 28.
Then, n(S) = 28.
Let E be the event of calling one of them by enroll number.
Given that, number of girls = 16.
Then, n(E) = 16.
The probability that the one called is a girl = n(S)/n(E) = 16/28 = 4/7.
Question 2
A box contains 3 white, 4 red and 7 blue erasers. If five erasers are taken at random then the probability that all the five are blue color is:
a) 13/211 b) 2/126 c) 3/286 d) 12/121
Answer : c) 3/286
Solution :
Total number of erasers in the box = 3 + 4 + 7 = 14.
Let S be the sample space.
Then, n(S) = number of ways of taking 5 out of 14.
Therefore, n(S) = 14C5 = (14 x 13 x 12 x 11 x 10)/(2 x 3 x 4 x 5) = 14 x 13 x 11 = 2002
Let E be the event of getting all the 5 blue erasers.
Therefore, n(E) = 7C5 = (7 x 6 x 5 x 4 x 3)/(2 x 3 x 4 x 5) = 21.
Now, the required probability = n(E)/n(S) = 21/2002 = 3/286.
Question 3
A box contains 6 bottles of variety1 drink, 3 bottles of variety2 drink and 4 bottles of variety3 drink. Three bottles of them are drawn at random, what is the
probability that the three are not of the same variety?
a) 833/858 b) 752/833 c) 632/713 d) none of these
Answer : a) 833/858
Solution :
Total number of drink bottles = 6 + 3 + 4 = 13.
Let S be the sample space.
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = 13C3 = (13 x 12 x 11)/(1 x 2 x 3) = 66 x 13 = 858.
Let E be the event of taking 3 bottles of the same variety.
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = 6C3 + 3C3 + 4C3
= 6 x 5 x 4 / 1 x 2 x 3 + 1 + 4 x 3 x 2 / 1 x 2 x 3
= 20 + 1 + 4 = 25.
The probability of taking 3 bottles of the same variety = n(E)/n(S) = 25/858.
Then, the probability of taking 3 bottles are not of the same variety = 1 - 25/858 = 833/858.
Question 4
There are 12 boys and 8 girls in a tution centre. If three of them scored first mark, then what is the probability that one of the three is a girl and the other two are boys?
a) 14/75 b) 22/55 c) 44/95 d) none of these
Answer : c) 44/95
Solution :
Total number of students = 20.
Let S be the sample space.
Then, n(S) = number of ways of three scored first mark
n(S) = 20C3 = 20 x 19 x 18 / 2 x 3 = 20 x 19 x 3
Let, E be the event of 1 girl and 2 boys.
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12.
n(E) = 8C1 x 12C2 = 8 x 12 x 11 / 1 x 2 = 8 x 6 x 11.
Now, the required probability = n(E)/n(S) = (8 x 6 x 11)/(20 x 19 x 3) = 44/95
