d MCQs on Probability Set-1 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

MCQs on Probability Set-1

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Q1. A problem of mathematics is given to 3 students, whose chances of solving it are 1/2,1/3, 1/4.Then the

probability that the problem is solved is

a. 4/13

b. 4/5

c. 3/4

d. 5/13

Answer: Option ‘c’ is correct



Let A, B  and C be the respective events of solving the problem.Then we are given that


P(A)= 1/2






Here clearly A,B and C are independent events and problem will stand resolved if any one of three students solve the problem.


Required probability is




Q2. Two dices are thrown. The chance or probability of getting an odd number on the first dice and a multiple of 3 on the other is

a. 1/5

b. 1/6

c. 1/4

d. none of these


Answer: option ‘b’ is correct




A= event of getting an odd number on the first dice

B= event of getting a multiple of 3 on second dice

We have





P(B)= 2/6=1/3

Since A and B are independent events;

Required probability is

P(A∩B)= P(A).P(B)=1/2 x 1/3 = 1/6


Q3. If a bag contains 3 red balls and 5 black balls and another bag contains 6 red balls and 4 black balls, then if one ball is drawn from both bags, then the probability that both will be red is;

a. 6/43

b. 9/40

c. 1/43

d. 8/41


Answer: Option ‘ b’ is correct




Let A be the event that ball drawn from first box is red and B be the event that the ball drawn


from second box is red.




P(A)= 3/8 and P(B)= 6/10


Required probability is;


P(A∩B)= P(A).P(B)= 3/8 x 6/10= 9/40

Q4. If P(A)=0.4, P(B)= p and P(AUB)=0.6 and given that events A and B are independent, then the value of p is;

a. 1/4

b. 1/2

c. 1/3

d. 1/5


Answer: Option ’ c’ is correct



As A nd B are given to be independent

⇒ P(A∩B)= P(A).P(B)

Also we know;



Put values;




⇒p= 0.2/0.6= 1/3

⇒ p= 1/3

Q5.If P(A̅)=0.7, P(B)=0.7 and P(B/A)=0.5 then P(A/B) is equal to

a. 3/14


c. 5/14

d. none of these

Answer: Option ’a’ is correct


Since P(A̅)=0.7



Now we know the formula;

P(B/A)= P(A∩B)/P(A)

⇒0.5= P(A∩B)/0.3

⇒P(A∩B)/= 0.15



⇒P(A/B)= 0.15/0.7


Q6. The odds in favor of an event are 3:5. The probability of occurrence of this event is;

a. 3/8

b. 8/11

c. 3/5

d. 7/11


Answer: Option ’a’ is correct



From given information odds are in favour ratio 3:5 i-e 3x are favorable cases and 5x are unfavorable cases.

So, total cases = 3x+5x=8x

So, probability of favorable cases is 3x/8x = 3/8

Q7. Out of 9 outstanding students in a college there are 4 boys and 5 girls. A team of  four students is to be selected for a quiz program. Then the probability that two are boys and two are girls is;


a. 12/31
b. 10/21
c. 10/31
d. 12/23
Answer: Option ’b; is correct
Possible ways that 4 students are selected is 9C4
Since there are 4 boys and 5 girls,  out out of which 2 boys and 2 girls are to be selected is possible ways;
Required probability= 4C2×5C2 / 9C4
= 10/21

Q8. A card is drawn from an ordinary pack of 52 cards and a gambler bets that it is a spade or an ace. Then the odds against his winning this bet is;
a. 8:5
b. 9:4
c. 9:5
d. 5:4
Answer: Option ’b’ is correct
We suppose ’A’ is the event of gettin a spade or an ace. So, possible number of cases for event ’A’ out of 52 cards is 52C1 =1
since thre are 13 cards for space including an ace for space and three aces other than an ace for spade.
Therefore favorable number of cases = 16C1=16
So, P(A)= 16/52=4/13
Hence odds against ’A’ are = P(A):P(A̅)= 9/13:4/13=9:4

Q9. A die is thrown. The probability of getting a prime number is;
a. 1/2
b. 1/4
c. 1/3
d. 1/5
Answer: Option ‘ a’ is correct
Let ‘S’ be the sample space for the experiment of throwing  a die i-e
S= {1,2,3,4,5,6}
Let ‘A’ be the event for getting a prime number
A= { 1,2,3,5}
P(A) = n(A)/n(S) = 3/6=1/2

Q10. A coin is tossed for three times successively. The probability of getting exactly one head or two heads is
a. 5/6
b. 4/5
c. 3/4
d. none of these
Answer: Option’ c’ is correct

Here let H= Head and T = tail

Sample space for the experiment is;


Let ’A’ be the event of getting one or two heads;



Number of elements in S i-e n(S)= 8

Number of elements in A i-e n(A)= 6

So, Probability for occurrence of event ’A’ is

P(A)= n(A)/n(S)=6/8=3/4

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