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

# MCQs on Probability Set-1

— math for CSS –, probability examples, probability & statistics, probability mcqs, probability examples and solutions

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

Explanation:

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

P(A)= 1/2

P(B)=1/3

P(C)=1/4

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

P(AUBUC)=1-P(A̅).P(B̅).P(C̅)=1-(1-1/2)(1-1/3)(1-1/4)=1-1/4=3/4

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

Explanation:

Let;

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

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

We have

A={1,3,5}

and

B={3,6}

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

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

Explanation:

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.

Then,

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

Explanation:

As A nd B are given to be independent

⇒ P(A∩B)= P(A).P(B)

Also we know;

P(A∪B)=P(A)+P(B)-P(A∩B)

⇒P(A∪B)=P(A)+P(B)-P(A).P(B)

Put values;

⇒0.6=0.4+p-0.4p

⇒0.6=0.4+p(1-0.4)

⇒0.2=0.6p

⇒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

b.3/5

c. 5/14

d. none of these

Explanation:

Since P(A̅)=0.7

⇒1-P(A)=0.7

P(A)=0.3

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

Again,

P(A/B)=P(A∩B)/P(B)

⇒P(A/B)= 0.15/0.7

⇒P(A/B)=3/14

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

Explanation:

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
Explanation:
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;
4C5C2
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
Explanation:
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
Explanation:
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
i-e
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
Explanation:

Here let H= Head and T = tail

Sample space for the experiment is;

S= {HHH,HHT,HTH,THH,HTT,TTH,THT,TTT}

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

A={HHT,HTH,THH,HTT,TTH,THT}

here,

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

•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•