1. There are 15 boys and 10 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
Explanation: Total number of ways of selecting 3 students from 25 students = 25C3
Number of ways of selecting 1 girl and 2 boys
= selecting 2 boys from 15 boys and 1 girl from 10 girls
⇒ Number of ways in which this can be done = 15C2 × 10C1
Ans: C
2. Three dice are thrown together. Find the probability of getting a total of at least 6.
Explanation: Since one die can be thrown in six ways to obtain any one of the six numbers marked on its six faces.
⇒ Total number of elementary events
= 6 × 6 × 6 = 216
Let A be the event of getting a total of at least 6. Then Ā denotes the event of getting a total of less than 6 i.e. 3, 4, 5
⇒ Ā = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1,2 ), (2, 2, 1)}
So, favourable number of cases = 10
Ans: C
3. A bag contains 6 pink and 5 yellow balls. One ball is drawn randomly. What is the probability that the ball drawn is pink?
4. In a box there are 10 apples and 2/5 th of the apples are rotten. If three apples are taken out from the box, what will be the probability that at least one apple is rotten?
5. A bag contains 5 yellow and 2 green and 3 red colour dice. If one dice from the bag are choosen at random, what is the probability that dice is either yellow or red colour?
6. A box contains 2 pink pens, 3 violet pens and 4 green pens. Find the probability of selecting 3 pens from the box such that at least 1 pen is green.
Explanation: Probability that at least 1 pen is green = 1 − Probability that none of the selected pens is green.
Now, number of ways in which no green pen is selected = 5C3 (as there are five non-green pens). And, number of ways of selecting three pens out of nine = 9C3 P(atleast 1 green pen).
And, number of ways of selecting three pens out of nine = 9C3
P(atleast 1 green pen)
Ans: D
7. There are 6 green, 5 red and 3 white balls in a bag. If 3 balls are drawn randomly what is the probability that no ball is red?
Ans: C
8. Probability of grasshopper eating grass = 1/5
Probability of frog eating grasshopper = 1/6
Probability of snake eating frog = 1/7
Probability of hawk eating snake = 1/8
Probability of man eating hawk = 1/9
What is the probability of a man eating a hawk who has eaten a snake which had consumed a frog who ate a grasshopper which didn’t eat grass?
Ans: B
9. There are 4 cotton kurtis, 3 woolen kurtis and 5 nylon kurtis. If 3 kurtis are selected at random, what is the probability that none of them are nylon kurtis?
Explanation: 3 kurtis out of 12 kurtis can be chosen in 12C3 ways.
As given in the question above that we don't have to choose any nylon kurti.
∴ We have to select 3 kurtis out of the remaining 7 kurtis.
Ans: C
10. There are 50 students in a class. 40% of the students like orange and 50% of the students like mango. If 10 students like both of them, then how many students like either orange or mango or both of them?
A) 30 B) 35 C) 40 D) 45 E) None of these
Explanation: The distribution of the fruits are given below:
The number students who like only mangoes: 40% of 50 = 20
The number students who like only oranges: 50% of 50 = 25
n(A U B) = n(A) + n(B) − n(A ∩ B)
Therefore, the number of students who like either orange or mango or both of them = 20 + 25 − 10 = 35
Ans: B
11. A box contains 21 balls numbered 1 to 21. A ball is drawn and then another ball is drawn without replacement. What is the probability that both balls are even numbered?
Explanation: There are 10 even numbers in the group 1 - 21.
∴ The probability that the first ball is even numbered = 10/21
Since the ball is not replaced there are now 20 balls left, of which 9 are even numbered.
∴ The probability that the second ball is even numbered = 9/20
∴ Required probability = 10/21 X 9/20 = 9/42 = 3/14
Ans: C
12. There are 3 green, 4 orange and 5 white color bulbs in a bag. If a bulb is picked at random, what is the probability of having either a green or a white bulb?
Explanation: Let E1, E2 be the event of picking a green bulb and white bulb respectively.
Total no. of bulbs in a bag = 3 + 4 + 5 = 12
Ans: B
13. A box contains slips with numbers from 1 to 50 written on them. A slip is drawn and replaced. Then another slip is drawn and after replacing another slip is drawn. What is the probability that an even number appears on the first draw, an odd number on the second draw and a number divisible by 3 on the third draw?
Explanation: The probability of an even number appearing on the first draw is 1/2 (since there are 25 even numbers in counting of 1 to 50). The probability of an odd number appearing on the second draw is 1/2 (since there are 25 odd numbers in counting of 1 to 50).
The probability of a number divisible by 3 appearing on the third draw is 16/50 (since there are 16 numbers that are divisible by 3 while counting from 1 to 50.)
Since all these events have no relation with each other and no dependence either, and the slips are replaced, we can directly multiply the individual probabilities to get the resultant probability. So, the probability of all the events taking place is
Ans: B
