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- Question
In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?
Options- A. 0.25
- B. 0.5
- C. 0.75
- D. 0.40
- Correct Answer
- 0.40
Explanation250 numbers between 101 and 350 i.e. n(S)=250
n(E)=100th digits of 2 = 299?199 = 100
P(E)= n(E)/n(S) = 100/250 = 0.40
- 1. You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?
Options- A. 1/2
- B. 1/12
- C. 2/3
- D. 3/4 Discuss
- 2. Two unbiased coin are tossed. What is the probability of getting atmost one head?
Options- A. 1/2
- B. 3/2
- C. 1/6
- D. 3/4 Discuss
- 3. Four persons are to be choosen from a group of 3 men, 2 women and 4 children. Find the probability of selecting 1 man,1 woman and 2 children.
Options- A. 2/7
- B. 3/7
- C. 4/7
- D. 3/7 Discuss
- 4. A number is selected from the numbers 1,2,3,4.......25.The probability for it to be divisible by 4 or 7 is:
Options- A. 3/25
- B. 9/25
- C. 1/25
- D. None of these Discuss
- 5. Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails
Options- A. 2
- B. 7/8
- C. 5/8
- D. 1/2 Discuss
- 6. Two dice are tossed. The probability that the total score is a prime number is :
Options- A. 1/6
- B. 1/2
- C. 5/12
- D. 3/4 Discuss
- 7. What is the probability of getting a sum 9 from two throws of a dice ?
Options- A. 1/6
- B. 1/2
- C. 1/9
- D. 3/4 Discuss
- 8. Let K and L be events on the same sample space, with P (K) = 0.8 and P (B) = 0.6. Are these two events are disjoint ?
Options- A. TRUE
- B. FALSE Discuss
- 9. A fair six-sided die is rolled twice. What is the probability of getting 4 on the first roll and not getting 6 on the second roll ?
Options- A. 1/36
- B. 5/36
- C. 1/12
- D. 1/9 Discuss
- 10. An unbiased die is rolled. Find the probability of getting a multiple of 3?
Options- A. 1/6
- B. 1/3
- C. 5/6
- D. None of these Discuss
Probability problems
Search Results
Correct Answer: 1/12
Explanation:
Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16
Required probability =(12)×(16)
= 1/12
Correct Answer: 3/4
Explanation:
Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4
Correct Answer: 2/7
Explanation:
Total number of persons = 9
Out of 9 persons 4 persons can be selected in ways =126
1 man,1 woman and 2 children can be selected in
ways =36
So,required probability = 36/126 =2/7
Correct Answer: 9/25
Explanation:
Total numbers = 25
Numbers divisible by 4 or 7 are 4, 7, 8, 12, 14, 16, 20, 21, 24 = 9
The probability (divisible by 4 or 7) = 9/25
Correct Answer: 1/2
Explanation:
S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }
=> n(S) = 8
E = { HHH, HHT, HTH, THH }
=> n(E) = 4
P(E) = 4/8 = 1/2
Correct Answer: 5/12
Correct Answer: 1/9
Explanation:
In two throws of a die, n(s)=(6 x 6)=36
let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)}
P(E) = n(E)/n(S) = 4/36 = 1/9
Correct Answer: FALSE
Explanation:
These two events cannot be disjoint because P(K) + P(L) > 1.
P(A?B) = P(A) + P(B) - P(A?B).
An event is disjoint if P(A ? B) = 0. If K and L are disjoint P(K ? L) = 0.8 + 0.6 = 1.4
And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.
Correct Answer: 5/36
Explanation:
The two events mentioned are independent.
The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.
P(getting first 4) = 1/6
P(no second 6) = 5/6
Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36
Correct Answer: 1/3
Explanation:
S = { 1, 2, 3, 4, 5, 6 }
=> n(S) = 6
E = { 3, 6}
=> n(E) = 2
Therefore, P(E) = 2/6 =1/3
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