Multiple Choice Questions On Aptitude | UGC NET Exam 2012 Sample

1)  A die is tossed once. What is the probability of getting an odd number?

1. 1/3
2. 1/2
3. 1
4. 3/4

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Answer = B 
Explanation:We know that the sample space for a die tossed once is { 1, 2, 3, 4, 5, 6} out of which {1, 3, 5} are odd numbers. So the p(getting an odd number)= (Favorable cases/Total cases) = 3/6 = 1/2

2) A coin is tossed once. Find the probability of getting a head ?

1. 1/2
2. 3/2
3. 3/4
4. None of above

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Answer = A 
Explanation:We know that sample space for a head tossed once is { head, tail} out of which head comes only once so the  P(getting head) = 1/2

3) A Bag contains 8 red, 7 white, 6 black balls. Find the probability of getting a black ball ?

1. 1
2. 6/8
3. 6/7
4. 2/7

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Answer =D 
Explanation:
Here Total number of balls are: 21
We have to find the probability of black ball and number of black balls are 6.
So the P(black ball) = 6/21 = 2/7
                                

4)  A bag contains 2 red, 8 black, and 7 white balls. Find the probability of not getting a white ball ?

1. 7/10
2. 7/8
3. 10/17
4. None

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Answer = C 
Explanation: 

Total balls=17
White balls = 7
P(white ball) = 7/17
P(Not white ball) = 1 - (7/17) = 10/17

5) Given that E and F are two events such that P(E) = 0.6, P(F) = 0.3 and P(E and F) = 0.2, Then what is the value of P(E/F)  ?

1. 2/3
2. 1/3
3. 1
4. None

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Answer = A 
Explanation: 

P(E/F) = P(E and F) / p(E) 
          = 0.2 / 0.3 
          = 2 / 3

6) A four digit number is formed, using the digits 1,2,3,5 with no repetitions. The probability that the number is divisible by 5 ?

1. 4/5
2. 1/4
3. 3/4
4. Can't say

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Answer = B 
Explanation: 

Let m be the favorable cases = Numbers of four digit which are divisible by 5 = 6
Let n be the total number of cases = ⁴C₄ = 24
Required Probability = 6/24 = 1/4

7) A coin is tossed twice. What is the probability that the head occurs at least once ?

1. 4/4
2. 2/4
3. 3/4
4. 0

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Answer = 3/4 
Explanation:

As a coin is tossed twice so sample space = {  HH , HT , TH , TT} Where H = head and T= tail
Sample space for favorable cases that the head occurs at least once = (HH, HT, TH}
Reqd. Probability = 3/4

8) A Coin is tossed twice. Find the probability that the head occurs at most once ?

1. 3/4
2. 1/4
3. 2/4
4. None

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Answer =  A
Explanation:

Sample space = {HH, HT, TH, TT}
Favorable cases = {HT, TH , TT}
P(Head occurs at most once) = 3/4 

9) Let E and F are two events. P(E) = 0.2, What is P(F) ?

1. can't be findable
2. 1
3. 2
4. 0.8

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Answer = D 
Explanation: 

As we know that the total probability of all the events = 1
P(One event)=0.2
P(second event)=1 - 0.2
                        = 0.8

10)  A die is thrown thrice. Then the total number of cases will be?

1. 3
2. 6
3. 8
4. 216

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Answer = D 
Explanation:
Total number of cases will be = 6 * 6  * 6 = 216

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