There are 15 boys and 10 girls in the class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
A. 1/40
B. 1/ 2
C. 21/46
D. 7/ 41
E. None of these
The correct option is: C
Solution:
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
⇒ Required probability = (15C2 × 10C1)/ (25C3)
The formula for combinations is nCk = n! / (k! * (n – k)!), where “!” denotes factorial.
Let’s calculate the combinations step by step:
- Calculate 15C2: 15C2 = 15! / (2! * (15 – 2)!) = 15! / (2! * 13!) = (15 * 14) / 2 = 210
- Calculate 10C1: 10C1 = 10! / (1! * (10 – 1)!) = 10! / (1! * 9!) = 10
- Calculate 25C3: 25C3 = 25! / (3! * (25 – 3)!) = 25! / (3! * 22!) = (25 * 24 * 23) / (3 * 2) = 2300
Now, let’s put it all together and solve the expression:
(15C2 × 10C1) / (25C3) = (210 × 10) / 2300 = 2100 / 2300 ≈ 0.91304348
Therefore, (15C2 × 10C1) / (25C3) is approximately 0.913.
Two friends Harish and Kalyan appeared for an exam. Let A be the event that Harish is selected and B is the event that Kalyan is selected. The probability of A is 2/5 and that of B is 3/7. Find the probability that both of them are selected.
A. 35/36
B. 5/35
C. 5/12
D. 6/35
E. None of these
The correct option is : D
Solution:
Given, A be the event that Harish is selected and
B is the event that Kalyan is selected.
P(A)= 2/5
P(B)=3/7
Let C be the event that both are selected.
P(C)=P(A)×P(B) as A and B are independent events:
P(C) = 2/5*3/7
P(C) =6/35
The probability that both of them are selected is 6/35
A card is drawn from a well-shuffled pack of 52 cards. What is the probability of getting a queen or club card?
A. 17/52
B. 15/52
C. 4/13
D. 3/13
E. None of these
The correct option is : C
Solution:
The probability of getting a queen card = 4/52
The probability of getting a club card = 13/52
The club card contains already a queen card, therefore the required probability is,
4/52 + 13/52 – 1/52 = 16/52 = 4/13
16 people shake hands with one another at a party. How many shaking hands took place?
A. 124
B. 120
C. 165
D. 150
E. None of these
The correct option is: B
Solution:
Total possible ways = 16C2
= 120
2 dice are thrown simultaneously. What is the probability that the sum of the numbers on the faces is divisible by either 3 or 5?
A. 7/36
B. 19/36
C. 9/36
D. 2/7
E. None of these
- The correct option is: B
Solution:
Clearly n(s) = 6*6 = 36
Let E be the event that the sum of the numbers on the 2 faces is divisible by either 3 or 5. Then
E = {(1,2), (1,4), (1,5), (2,1), (2,3), (2,4), (3,2), (3,3), (3,6), (4,1), (4,2), (4,5), (4,6), (5,1), (5,4), (5,5), (6,3), (6,4), (6,6)}
n(E) = 19
Hence P(E) = n(E) / n(s)
= 19/ 36
Daniel speaks the truth in 2/5 cases and Sherin lies in 3/7 cases. What is the percentage of cases in which both Daniel and Sherin contradict each other in stating a fact?
A. 72.6%
B. 51.4%
C. 62.3%
D. 47.5%
E. None of these
The correct option is: B
Solution:
Daniel and Sherin will contradict each other when one speaks the truth and the other speaks lies.
The probability of Daniel speaking the truth and Sherin lies
=2/5*3/7
=6/35
The probability of Sherin speaking the truth and Daniel lies
=4/7*3/5
=12/35
The two probabilities are mutually exclusive.
Hence, probabilities that Daniel and Sherin contradict each other:
=6/35 +12/35
=18/35
=18/35*100
=51.4%
The names of 5 students from Section A, 6 students from Section B and 7 students from Section C were selected. The age of all 18 students was different. Again, one name was selected from them and it was found that it was of section B. What was the probability that it was the youngest student of the section B?
A. 1/18
B. 1/15
C. 1/6
D. 1/12
E. None of these
The correct option is: C
Solution:
The total number of students = 18
When 1 name was selected from 18 names, the probability that he was of section B
= 6 = 1
18 3
But from the question, there are 6 students from section B and the age of all 6 are different therefore, the probability of selecting one i.e. youngest student from 6 students will be 1/6
There are a total of 18 balls in a bag. Out of them 6 are red in colour, 4 are green in colour and 8 are blue in colour. If Vishal picks three balls randomly from the bag, then what will be the probability that all three balls are not of the same colour?
A. 95/102
B. 19/23
C. 21/26
D. 46/51
E. None of these
The correct option is : D
Solution:
Number of ways in which the person can pick three balls out of 18 balls
= 18C3 = 816
Number of ways of picking 3 balls of same colour = 6C3+ 4C3 + 8C3 = (20 + 4 + 56) = 80
Probability of picking three balls of the same color
= 80 = 5
816 51
Required probability = 1 – the probability of picking three balls of the same colour
= 1 – 5 = 46
Bag A contains 3 green and 7 blue balls. While bag B contains 10 green and 5 blue balls. If one ball is drawn from each bag, what is the probability that both are green?
A. 29/30
B. 1/5
C. 1/3
D. 1/30
E. None of these
- The correct option is: B
Solution:
The required probability = 3C1/10C1 × 10C1/15C1
= 3/10 × 10/15 = 1/5
Ram and Shyam are playing chess together. Ram knows the two rows in which he has to put all the pieces in but he doesn’t know how to place them. What is the probability that he puts all the pieces in the right place?
A. 8!/16!
B. 8!/(2*15!)
C. 8!/15!
D. (2*8!)/16!
E. None of these
The correct option is: B
Solution:
Total boxes = 16
Total pieces = 16
Similar pieces = 8 pawns, 2 bishops, 2 rooks, 2 knights
Total ways of arranging these 16 pieces in 16 boxes
= 16! = 16!
(8! 2! 2! 2!) (8 × 8!)
Ways of correct arrangement = 1
Probability of correct arrangement = 1
(16! / (8 × 8!)
= (8 × 8!) = 8!
16! (2 × 15!)
To solve the expression 16! / (2 × 15!), we can simplify it step by step:
- Calculate 15!: 15! = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
- Multiply 2 by 15!: 2 × 15! = 2 × (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)
- Simplify 16! / (2 × 15!) = 16! / (2 × (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1))
Now, let’s look at the factorial expressions:
- 16! = 16 × 15! (because 16! = 16 × 15!),
- 15! = 15 × 14! (because 15! = 15 × 14!),
- 14! = 14 × 13! (because 14! = 14 × 13!), and so on until 2! = 2 × 1 (because 2! = 2 × 1).
We can now rewrite the expression as:
16! / (2 × 15!) = (16 × 15!) / (2 × (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1))
Cancel out the common factors:
16! / (2 × 15!) = (16 × 15!) / (2 × 15!) = 16 / 2 = 8
So, 16! / (2 × 15!) equals 8.
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