# 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**

**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**

**Answer: option ‘b’ is correct**

**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**

**Answer: Option ’a’ is correct**

**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**

**Answer: Option ’a’ is correct**

**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**

**Answer: Option ’b; is correct**

**Explanation:**

^{9}C

_{4}

_{Since there are 4 boys and 5 girls, out out of which 2 boys and 2 girls are to be selected is possible ways;}

^{4}C

_{2×}

^{5}C

_{2}

_{ }

_{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**

**Explanation:**

^{52}C

_{1 }=1

^{16}C

_{1}=16

**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:**

**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**

**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**