# Probability Questions and Answers updated daily – Aptitude

Probability Questions: Solved 93 Probability Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Probability Questions

1. A committee of 3 members is to be made out of 6 men and 5 women. What is the probability that the committee has at least two women?

SHOW ANSWER

Correct Ans:(14/33)

Explanation:

Number of possible combination of 3 persons in which 2 have to be women = (2 Women out of 5 * 1 Man out of 6) or (3 Women out of 5)

= (

Total possible outcomes =

= ( 5!/2! * 3! * 6!/5! * 1!+ 5!/3! * 2!)/(11!/3! * 8!) = 70/11 * 15 =

= (

^{5}C_{2}*^{6}C_{1}+^{5}C_{3})Total possible outcomes =

^{11}C_{3}= ( 5!/2! * 3! * 6!/5! * 1!+ 5!/3! * 2!)/(11!/3! * 8!) = 70/11 * 15 =

**(14/33)****Hence, option B is correct.**
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2. There are 3 green, 4 orange and 5 white color bulbs in a bag. If a bulb is picked at random, what is the probability of having either a green or a white bulb?

SHOW ANSWER

Correct Ans:(2/3)

Explanation:

Let E1, E2 be the event of picking a green bulb and white bulb respectively.

Total no. of bulbs in a bag = 3 + 4 + 5 = 12

E1 = 3/12 = 1/4

E2 = 5/12 = 5/12

P(E1 or E2) = P(E1) + P(E2)

= 1/4 + 5/12 = (2/3)

Total no. of bulbs in a bag = 3 + 4 + 5 = 12

E1 = 3/12 = 1/4

E2 = 5/12 = 5/12

P(E1 or E2) = P(E1) + P(E2)

= 1/4 + 5/12 = (2/3)

**Hence, option B is correct.**
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3. A box contains 21 balls numbered 1 to 21. A ball is drawn and then another ball is drawn without replacement. What is the probability that both balls are even numbered?

SHOW ANSWER

Correct Ans:(3/14)

Explanation:

There are 10 even numbers in the group 1-21.

The probability that the first ball is even numbered = 10/21

Since the ball is not replaced there are now 20 balls left, of which 9 are even numbered.

The probability that the second ball is even numbered = 9/20

Required probability = 10/21 * 9/20 = 9/42 = 3/14

The probability that the first ball is even numbered = 10/21

Since the ball is not replaced there are now 20 balls left, of which 9 are even numbered.

The probability that the second ball is even numbered = 9/20

Required probability = 10/21 * 9/20 = 9/42 = 3/14

**Hence, option C is correct.**
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4. There are three events X,Y and Z, one of which must and only can happen. If the odds are 7:4 against X, 5:3 against Y, the odds against Z must be:

SHOW ANSWER

Correct Ans:(65/23)

Explanation:

According to the question,

P(Xâ€™)/P(X) = (7/4)

P(Xâ€™) = (7/11)

P(X) = (4/11)

P(Yâ€™)/P(Y) = (5/3)

P(Yâ€™) = (5/8) ,P(Y) = (3/8)

Now, out of X,Y and Z, one and only one can happen.

P(X)+P(Y)+P(Z) = 1

(4/11) + (3/8) + P(Z) = 1

P(Z)= 1 - (4/11) - (3/8)

= 88 - 32 - (33/88)

= (23/88)

P(Zâ€™) = 1 - P(Z)

= 1 - (23/88)

=

So odd against z

P(zâ€™)/p(z) =

P(Xâ€™)/P(X) = (7/4)

P(Xâ€™) = (7/11)

P(X) = (4/11)

P(Yâ€™)/P(Y) = (5/3)

P(Yâ€™) = (5/8) ,P(Y) = (3/8)

Now, out of X,Y and Z, one and only one can happen.

P(X)+P(Y)+P(Z) = 1

(4/11) + (3/8) + P(Z) = 1

P(Z)= 1 - (4/11) - (3/8)

= 88 - 32 - (33/88)

= (23/88)

P(Zâ€™) = 1 - P(Z)

= 1 - (23/88)

=

**(65/88)**So odd against z

P(zâ€™)/p(z) =

**(65/23)**
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5. The probability that a bullet fired from a point will strike the target is 3/4. 5 such bullets are fired simultaneously towards the target from that very point. What is the probability that the target will be strike?

SHOW ANSWER

Correct Ans:(255/256)

Explanation:

Probability of a bullet not striking the target = Â¼

Probability that none of the 5 bullets will strike the target

= (1/4)

= (1/256)

Probability that the target will strike at least once:

= 1- (1/256)

=

Probability that none of the 5 bullets will strike the target

= (1/4)

^{5}= (1/256)

Probability that the target will strike at least once:

= 1- (1/256)

=

**(255/256)**
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6. Preethi was born between August 26

^{th}and 30^{th}(26^{th}and 30^{th}excluding). Her year of birth is also unknown. What is the probability of Preethi being born on a Monday?SHOW ANSWER

Correct Ans:(3/7)

Explanation:

Since the year of birth is Unknown, the birthday being on Monday can have a zero probability.

Also since between 26

i.e. 27

We have following possibilities on these three dates,

Monday, Tuesday, Wednesday

Tuesday, Wednesday, Thursday

Wednesday, Thursday, Friday

Thursday, Friday, Saturday,

Friday, Saturday, Sunday

Saturday, Sunday, Monday

Sunday, Monday, Tuesday

Out of these 7 events, we have 3 chances of his birthday falling on Monday

Probability = favourable events/ total events

Probability = (3/7)

Therefore, the probability of birthday falling on Monday can be

Also since between 26

^{th}and 30^{th}, there are three daysi.e. 27

^{th}, 28^{th}and 29^{th}We have following possibilities on these three dates,

Monday, Tuesday, Wednesday

Tuesday, Wednesday, Thursday

Wednesday, Thursday, Friday

Thursday, Friday, Saturday,

Friday, Saturday, Sunday

Saturday, Sunday, Monday

Sunday, Monday, Tuesday

Out of these 7 events, we have 3 chances of his birthday falling on Monday

Probability = favourable events/ total events

Probability = (3/7)

Therefore, the probability of birthday falling on Monday can be

**(3/7)**.
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7. A box contains 6 pink, 4 red and 5 yellow marbles. If 2 marbles are picked at random, then find the probability that the marbles are either yellow or pink?

SHOW ANSWER

Correct Ans:5/21

Explanation:

Total probability = 15C

Required probability = 5C

The probability that the marbles are either red or pink,

= 5C

= (5*4/2 + 6*5/2) / (15*14/2)

= (5*2 + 3*5) / (15*7)

= 25/105

= 5/21

_{2}Required probability = 5C

_{2}or 6C_{2}The probability that the marbles are either red or pink,

= 5C

_{2}or 6C_{2}/ 15C_{2}= (5*4/2 + 6*5/2) / (15*14/2)

= (5*2 + 3*5) / (15*7)

= 25/105

= 5/21

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8. Probability of taking alternate balls of red and blue balls is 308/4845. 4 balls are drawn one by one and not replaced. If there are 12 red balls and total number of balls is 20, which of the following can be found?

A) Probability of taking both red balls

B) Probability of taking one blue ball

C) Probability of taking 1 ball

A) Probability of taking both red balls

B) Probability of taking one blue ball

C) Probability of taking 1 ball

SHOW ANSWER

Correct Ans:None of these

Explanation:

308/4845 = (12 / 20) * (y / 19) * (11 / 18) * ((y - 1) / 17))

308/4845 = 11y(y - 1) / (10*19*3*17)

28*2 = y(y - 1)

y

y = -7, 8

y = 8

So, 8 blue balls, A, B, C can be found.

308/4845 = 11y(y - 1) / (10*19*3*17)

28*2 = y(y - 1)

y

^{2}- y - 56 = 0y = -7, 8

y = 8

So, 8 blue balls, A, B, C can be found.

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9. A box consists of 15 balls numbered from 0 to 14. A boy picked a ball from the box and kept it in the bag after noting its number. He repeated this process 2 more times. What is the probability that the ball picked first by the boy is numbered higher than the ball picked second and the ball picked second by the boy is numbered higher than the ball picked third?

SHOW ANSWER

Correct Ans:2197/3375

Explanation:

Let the number on the ball picked first = a, second = b, third = c.

The order of a, b and c can be (a > b > c)

Number of ways selecting the first number = 13 (Because we canâ€™t select 0 and 1)

Number of ways selecting the second number = 13 (Because we canâ€™t select 0)

Number of ways selecting the third number = 13 (Because we canâ€™t select first and third number)

Required probability = (13 * 13 * 13)/15

= 2197/3375

The order of a, b and c can be (a > b > c)

Number of ways selecting the first number = 13 (Because we canâ€™t select 0 and 1)

Number of ways selecting the second number = 13 (Because we canâ€™t select 0)

Number of ways selecting the third number = 13 (Because we canâ€™t select first and third number)

Required probability = (13 * 13 * 13)/15

^{3}= 2197/3375

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10. Bag A contains 'P' green and 18 yellow balls while bag B contains '(P + 2)' green balls and 22 more number of yellow balls than that of in bag A. Probability of selecting a green ball from bag A is 1/12 more than probability of selecting a green ball from bag B. Find total number of balls in bag B. (P < 50)

SHOW ANSWER

Correct Ans:48

Explanation:

From the given data,

No. of green balls = P

No. of yellow balls = 18

No. of green balls = P + 2

No. of yellow balls = 18 + 22 = 40

Now, Probability of selecting a green ball from bag A = No. of green balls in bag A/ Total no. of balls in bag A

= P/(P + 18)

Probability of selecting a green ball from bag B = (P + 2)/(P + 2 + 40)

= (P + 2)/(P + 42)

Given that, Probability of selecting a green ball from bag A = (1/12) + Probability of selecting a green ball from bag B

---> P/(P + 18) = (1/12) + [(P + 2)/(P + 42)]

---> [P/(P + 18)] - [(P + 2)/(P + 42)] = 1/12

---> [P(P + 42) - (P + 2)(P + 18)] / (P + 18)(P + 42) = 1/12

---> [P

---> [P

---> [ 42P - 20P - 36] / (P

---> [22P - 36]/(P

---> 12[22P - 36] = (P

---> 264P - 432 = P

---> P

---> P

We can find the value of P by using the formula, P = [-b ± √(b

Where, a = 1

b = -204

c = 1188

---> P = [204 ± √((-204)

= [204 ± √(41616 - 4752)] / 2

= [204 ± √(36864â€¬)] / 2

= [204 ± 192] / 2

= [204 + 192]/2; [204 - 192]/2

= 396/2; 12/2

= 198; 6

Thus,

Given that, P < 50, So P canâ€™t be 198.

Hence,

Now,

= 6 + 2 + 40

=

**In bag A:**No. of green balls = P

No. of yellow balls = 18

**In bag B:**No. of green balls = P + 2

No. of yellow balls = 18 + 22 = 40

Now, Probability of selecting a green ball from bag A = No. of green balls in bag A/ Total no. of balls in bag A

= P/(P + 18)

Probability of selecting a green ball from bag B = (P + 2)/(P + 2 + 40)

= (P + 2)/(P + 42)

Given that, Probability of selecting a green ball from bag A = (1/12) + Probability of selecting a green ball from bag B

---> P/(P + 18) = (1/12) + [(P + 2)/(P + 42)]

---> [P/(P + 18)] - [(P + 2)/(P + 42)] = 1/12

---> [P(P + 42) - (P + 2)(P + 18)] / (P + 18)(P + 42) = 1/12

---> [P

^{2}+ 42P - (P^{2}+ 18P + 2P + 36)] / (P^{2}+ 42P + 18P + 756) = 1/12---> [P

^{2}+ 42P - P^{2}- 20P - 36] / (P^{2}+ 60P + 756) = 1/12---> [ 42P - 20P - 36] / (P

^{2}+ 60P + 756) = 1/12---> [22P - 36]/(P

^{2}+ 60P + 756) = 1/12---> 12[22P - 36] = (P

^{2}+ 60P + 756)---> 264P - 432 = P

^{2}+ 60P + 756---> P

^{2}+ 60P - 264P + 756 + 432 = 0---> P

^{2}- 204P + 1188 = 0We can find the value of P by using the formula, P = [-b ± √(b

^{2}- 4ac)] / 2aWhere, a = 1

b = -204

c = 1188

---> P = [204 ± √((-204)

^{2}- 4 * 1188)] / 2= [204 ± √(41616 - 4752)] / 2

= [204 ± √(36864â€¬)] / 2

= [204 ± 192] / 2

= [204 + 192]/2; [204 - 192]/2

= 396/2; 12/2

= 198; 6

Thus,

**P = 198 or 6**Given that, P < 50, So P canâ€™t be 198.

Hence,

**P = 6**Now,

**Total number of balls in bag B**= P + 2 + 40= 6 + 2 + 40

=

**48**
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11. The probability that a student will pass in Mathematics is 3/5, and the probability that he will pass in English is 1/3. If the probability that he will pass in both Mathematics and English is 1/8, what is the probability that he will pass in at least one subject?

SHOW ANSWER

Correct Ans:(79/120)

Explanation:

Given, Probability that a student will pass in Mathematics = 3/5

---> Probability that a student will fail in Mathematics = 1 - (3/5) = 2/5

Given, Probability that a student will pass in English = 1/3

---> Probability that a student will fail in English = 1 - (1/3) = 2/3

Given to find the probability that he will pass in at least one subject. Hence, there will be 3 cases:

Case 1: The student will pass in Mathematics and fail in English

Case 2: The student will pass in English and fail in Mathematics

Case 3: The student will pass in both Mathematics and English

Now, P(M) = Probability that a student will pass in Mathematics and failed in English

= (3/5) x (2/3)

= 2/5

P(E) = Probability that a student will pass in English and failed in Mathematics

= (1/3) x (2/5)

= 2/15

P(M ∩ E) = Probability that a student will pass in both Mathematics and English

= 1/8

= (2/5) + (2/15) + (1/8)

= (48 + 16 + 15)/120

=

---> Probability that a student will fail in Mathematics = 1 - (3/5) = 2/5

Given, Probability that a student will pass in English = 1/3

---> Probability that a student will fail in English = 1 - (1/3) = 2/3

Given to find the probability that he will pass in at least one subject. Hence, there will be 3 cases:

Case 1: The student will pass in Mathematics and fail in English

Case 2: The student will pass in English and fail in Mathematics

Case 3: The student will pass in both Mathematics and English

Now, P(M) = Probability that a student will pass in Mathematics and failed in English

= (3/5) x (2/3)

= 2/5

P(E) = Probability that a student will pass in English and failed in Mathematics

= (1/3) x (2/5)

= 2/15

P(M ∩ E) = Probability that a student will pass in both Mathematics and English

= 1/8

**Required Probability**= P(M) + P(E) + P(M ∩ E)= (2/5) + (2/15) + (1/8)

= (48 + 16 + 15)/120

=

**79/120**
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12. Fourteen applications apply for a job out of which there are 9 men and 5 women. If only three applications selected for the job then find the probability that at least one of the selected application is of a woman?

SHOW ANSWER

Correct Ans:(10/13)

Explanation:

Total number of ways to select three persons =

= (14 * 13 * 12)/(3 * 2 * 1)

= 14 * 13 * 2

= 364

Given condition, atleast one woman should be there out of 3 selected applications.

So, Required case = 2 men 1 woman (or) 1 man 2 women (or) 3 women

= (

= (36 * 5) + (9 * 10) + (10)

= 180 + 90 + 10

= 280

^{14}C_{3}= (14 * 13 * 12)/(3 * 2 * 1)

= 14 * 13 * 2

= 364

Given condition, atleast one woman should be there out of 3 selected applications.

So, Required case = 2 men 1 woman (or) 1 man 2 women (or) 3 women

= (

^{9}C_{2}*^{5}C_{1}) + (^{9}C_{1}*^{5}C_{2}) + (^{5}C_{3})= (36 * 5) + (9 * 10) + (10)

= 180 + 90 + 10

= 280

**Required Probability**= 280/364 =**10/13**
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13. Akshaya and Nikitha play a game where each is asked to select a number from 1 to 8. If the two numbers same, both of them win a prize. The probability that they will not win a prize in a single trial is?

SHOW ANSWER

Correct Ans:(7/8)

Explanation:

----> Total number of ways in which both of them can select a number each:

----> = (8*8)

----> = 64

---->Total number of ways in which both of them can select a same number so that they both can win:

----> = 8ways

---->[They both can select {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8)}]

---->Probability that they win the prize:

----> = (Favourable Cases / Total Cases)

----> = (8/64)

----> = (1/8)

----> Probability that they do not win a prize:

----> = (1-(1/8))

----> = (7/8)

----> = (8*8)

----> = 64

---->Total number of ways in which both of them can select a same number so that they both can win:

----> = 8ways

---->[They both can select {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8)}]

---->Probability that they win the prize:

----> = (Favourable Cases / Total Cases)

----> = (8/64)

----> = (1/8)

----> Probability that they do not win a prize:

----> = (1-(1/8))

----> = (7/8)

**Hence the answer is : (7/8)**
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14. Two friends Harish and Kalyan appeared for an exam. Let A be the event that Harish is selected and B is the event that Kalyan is selected. The probability of A is (2/5) and that of B is (3/7). Find the probability that both of them are selected.

SHOW ANSWER

Correct Ans:(6/35)

Explanation:

----> Given, A be the event that Harish is selected

----> B is the event that Kalyan is selected.

----> P(A) = (2/5)

----> P(B) = (3/7)

----> Let C be the event that both are selected.

----> P(C) = P(A) * P(B) as A and B are independent events:

----> P(C) = (2/5) * (3/7)

----> P(C) = (6/35)

----> The probability that both of them are selected is (6/35)

----> B is the event that Kalyan is selected.

----> P(A) = (2/5)

----> P(B) = (3/7)

----> Let C be the event that both are selected.

----> P(C) = P(A) * P(B) as A and B are independent events:

----> P(C) = (2/5) * (3/7)

----> P(C) = (6/35)

----> The probability that both of them are selected is (6/35)

**Hence the answer is : (6/35)**
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15. Srinaya forgot the last digit of an 11 digit landline number. If she randomly dials the final 2 digits after correctly dialing the first nine, then what is the chance of dialing the correct number?

SHOW ANSWER

Correct Ans:(1/100)

Explanation:

Srinaya forgot the last digit of an 11 digit landline number. If she randomly dials the final 2 digits after correctly dialing the first nine, then what is the chance of dialing the correct number:

Reference:

----> It is given that last two digits are randomly dialed.

----> Then each of the digits can be selected out of 10 digits in 10 ways.

----> Hence required probability

----> = (1/10)2

----> = (1/100)

Reference:

----> It is given that last two digits are randomly dialed.

----> Then each of the digits can be selected out of 10 digits in 10 ways.

----> Hence required probability

----> = (1/10)2

----> = (1/100)

**Hence the answer is : (1/100)**
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16. A bag "A" contains 4 black & 6 red balls and bag "B" contains 7 black & 3 red balls. A die is thrown if 1 or 2 appears on it, then bag A is chosen , otherwise bag B chosen . If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black.

SHOW ANSWER

Correct Ans:11/45

Explanation:

If a die is thrown, the probability of 1 or 2 appearing = 2/6 = 1/3

Hence, the probability of any other number appearing = 1 - 1/3 = 2/3

In bag 'A' the probability of one red and one black = (6*4)/(10*9) = 12/45

In bag 'B' the probability of one red and one black = (3*7)/(10*9) = 7/30

So the probabilITY of one red and another black ball = 1/3 * 12/45 + 2/3 * 7/30

= 4/45 + 7/45 = 11/45

Hence, the probability of any other number appearing = 1 - 1/3 = 2/3

In bag 'A' the probability of one red and one black = (6*4)/(10*9) = 12/45

In bag 'B' the probability of one red and one black = (3*7)/(10*9) = 7/30

So the probabilITY of one red and another black ball = 1/3 * 12/45 + 2/3 * 7/30

= 4/45 + 7/45 = 11/45

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17. There are 50 students in a class. 40% of the students like Orange and 50% of the students like Mango. If 10 students like both of them, then how many students like either Orange or Mango or both of them?

SHOW ANSWER

Correct Ans:35

Explanation:

Given, Total students = 50

The number of students who like only Oranges = 40% of 50 = 20

The number of students who like only Mangoes = 50% of 50 = 25

Number of students who like both Oranges and Mangoes = 10

W.K.T:

Therefore, the number of students who like either Orange or Mango or both of them = 20 + 25 - 10 = 35

The number of students who like only Oranges = 40% of 50 = 20

The number of students who like only Mangoes = 50% of 50 = 25

Number of students who like both Oranges and Mangoes = 10

W.K.T:

**n(A ∪ B) = n(A) + n(B) - n(A ∩ B)**Therefore, the number of students who like either Orange or Mango or both of them = 20 + 25 - 10 = 35

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18. A bag contains red, black, white and brown balls. The probability of getting a red ball from a bag filled with balls is 2/13 and the number of black balls in the bag is 5. If the white ball is 30% less than the brown ball and 40% more than the black ball, find the number of red balls.

SHOW ANSWER

Correct Ans:4

Explanation:

Let the number of red balls = 2x

Total balls = 13x

Now ATQ

Black balls = 5

White balls = 5 * 140/100 = 7

Brown balls = 7 * 100/70 = 10

So,

2x + 5 + 7 + 10 = 13x

22 = 11x

x = 2

Red balls = 2*2 = 4

Total balls = 13x

Now ATQ

Black balls = 5

White balls = 5 * 140/100 = 7

Brown balls = 7 * 100/70 = 10

So,

2x + 5 + 7 + 10 = 13x

22 = 11x

x = 2

Red balls = 2*2 = 4

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19. A bag contains 5 black and 7 white balls. A ball is drawn out-of it and replaced in the bag. Then a ball is drawn again. What is the probability that (i) both the balls drawn were black, (ii) the first ball was black and the second white?

SHOW ANSWER

Correct Ans:(25/144), (35/144)

Explanation:

The events are independent and capable of simultaneous occurence. The rule of multiplication would be applied.

The Probability that,

= (5/12) * (5/12) ----> (i.e., Probability of drawing black ball in first draw as well as second draw)

=

= (5/12) * (7/12)

=

The Probability that,

**(i) Both the balls drawn were black**= (5/12) * (5/12) ----> (i.e., Probability of drawing black ball in first draw as well as second draw)

=

**25/144****(ii) the first ball was black and the second white**= (5/12) * (7/12)

=

**35/144**
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20. A box contains slips with numbers from 1 to 50 written on them. A slip is drawn and replaced. Then another slip is drawn and after replacing another slip is drawn. What is the probability that an even number appears on the first draw, an odd number on the second draw and a number divisible by 3 on the third draw?

SHOW ANSWER

Correct Ans:(2/25)

Explanation:

The probability of an even number appearing on the first draw = 25/50

(---> since there are 25 even numbers in counting of 1 to 50)

= 1/2

The probability of an odd number appearing on the second draw = 25/50

(---> since there are 25 odd numbers in counting of 1 to 50)

= 1/2

The probability of a number divisible by 3 appearing on the third draw = 16/50

(---> since there are 16 numbers that are divisible by 3 while counting from 1 to 50)

Since all these events have no relation with each other and no dependence either, and the slips are replaced, we can directly multiply the individual probabilities to get the resultant probability.

So, the

= 16/200

=

(---> since there are 25 even numbers in counting of 1 to 50)

= 1/2

The probability of an odd number appearing on the second draw = 25/50

(---> since there are 25 odd numbers in counting of 1 to 50)

= 1/2

The probability of a number divisible by 3 appearing on the third draw = 16/50

(---> since there are 16 numbers that are divisible by 3 while counting from 1 to 50)

Since all these events have no relation with each other and no dependence either, and the slips are replaced, we can directly multiply the individual probabilities to get the resultant probability.

So, the

**required probability**= (1/2) * (1/2) * 16/50)= 16/200

=

**2/25**
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