18 + Interview Questions in QUANTITATIVE APTITUDE in PROBABILITY Page 1 InterviewSolution

1.	Quantity 2: A bag contains 25 balls which are red, green and blue. The probability of picking a red ball first is 2/5. If the first ball was red and without keeping it back, probability of picking a blue ball is 1/3, find the number of green balls.1). Quantity 1 > Quantity 22). Quantity 1 < Quantity 23). Quantity 1 ≥ Quantity 24). Quantity 1 ≤ Quantity 2
Answer» Solving for Quantity 1: Probability of picking a blue ball = no. of blue balls/20 = 2/5 ⇒ No. of blue balls = 40/5 = 8 Probability of picking EITHER a red or blue ball = (no. of red balls + no. of blue balls) /20 = ¾ ⇒ No. of red balls + no. of blue balls = 15 ⇒ No. of red balls = 15 – 8 = 7 ⇒ Quantity 1 = 7 Solving for Quantity 2: Probability of picking a red ball first = no. of red balls/25 = 2/5 ⇒ No. of red balls = 10 Probability of picking a blue ball without keeping the first ball BACK = no. of blue balls/24 = 1/3 ⇒ No. of blue balls = 8 ⇒ No. of green balls = 25 – (10 + 8) = 25 – 18 = 7 ⇒ Quantity 2 = 7 ∴ Quantity 1 = Quantity 2

1.

Quantity 2: A bag contains 25 balls which are red, green and blue. The probability of picking a red ball first is 2/5. If the first ball was red and without keeping it back, probability of picking a blue ball is 1/3, find the number of green balls.1). Quantity 1 > Quantity 22). Quantity 1 < Quantity 23). Quantity 1 ≥ Quantity 24). Quantity 1 ≤ Quantity 2

Answer»

Solving for Quantity 1:

Probability of picking a blue ball = no. of blue balls/20 = 2/5

⇒ No. of blue balls = 40/5 = 8

Probability of picking EITHER a red or blue ball = (no. of red balls + no. of blue balls) /20 = ¾

⇒ No. of red balls + no. of blue balls = 15

⇒ No. of red balls = 15 – 8 = 7

⇒ Quantity 1 = 7

Solving for Quantity 2:

Probability of picking a red ball first = no. of red balls/25 = 2/5

⇒ No. of red balls = 10

Probability of picking a blue ball without keeping the first ball BACK = no. of blue balls/24 = 1/3

⇒ No. of blue balls = 8

⇒ No. of green balls = 25 – (10 + 8) = 25 – 18 = 7

⇒ Quantity 2 = 7

∴ Quantity 1 = Quantity 2

Discussion

2.	Rohan flips a coin, rolls a dice and draws a card. What is the probability of getting a head, 4 on dice and a card of king together?1). Chances are more than 1%2). Chances are less than 0.10%3). Chances are more than 2%4). Chances are more than 0.50%
Answer» Probability of GETTING a HEAD on COIN = ½ Probability of getting a 4 on DICE = 1/6 Probability of getting a card of KING = 4/52 = 1/13 ⇒ Probability of getting a head, 4 on dice and a card of king together = ½ × 1/6 × 1/13 = 0.0064 ⇒ 0.64% ∴ Chances are more than 0.50%

2.

Rohan flips a coin, rolls a dice and draws a card. What is the probability of getting a head, 4 on dice and a card of king together?1). Chances are more than 1%2). Chances are less than 0.10%3). Chances are more than 2%4). Chances are more than 0.50%

Answer»

Probability of GETTING a HEAD on COIN = ½

Probability of getting a 4 on DICE = 1/6

Probability of getting a card of KING = 4/52 = 1/13

⇒ Probability of getting a head, 4 on dice and a card of king together = ½ × 1/6 × 1/13 = 0.0064 ⇒ 0.64%

∴ Chances are more than 0.50%

Discussion

3.	Three packs of cards are thoroughly mixed and shuffled and 2 cards are drawn at random, one after the other. What is the probability that all of them are jacks?1). 43/23052). 4/33553). 7/13354). 98/1695
Answer» No. of CARDS = 52 × 3 = 156 No. of jacks = 4 × 3 = 12 Number of ways of DRAWING 2 jacks out of total 12 jacks = ¹²C₂ Number of ways of drawing 2 cards out of 156 = ¹⁵⁶C₂ Required probability = ¹²C₂/¹⁵⁶C₂ = (12 × 11)/(156 × 155) = 11/2015

3.

Three packs of cards are thoroughly mixed and shuffled and 2 cards are drawn at random, one after the other. What is the probability that all of them are jacks?1). 43/23052). 4/33553). 7/13354). 98/1695

Answer»

No. of CARDS = 52 × 3 = 156

No. of jacks = 4 × 3 = 12

Number of ways of DRAWING 2 jacks out of total 12 jacks = ¹²C₂

Number of ways of drawing 2 cards out of 156 = ¹⁵⁶C₂

Required probability = ¹²C₂/¹⁵⁶C₂ = (12 × 11)/(156 × 155) = 11/2015

Discussion

4.	In a class consisting of 5 girls and 6 boys, they are to be seated on 11 chairs in a row in such a way that all the boys always sit together. In how many different ways can this be done?1). 4320002). 4800003). 5280004). 518400
Answer» Consider the 6 BOYS as one single unit The total number of people will be 5 (girls) + 1 (All boys considered as one unit) = 6 These 6 people can be ARRANGED in 6! WAYS = 720 ways The 6 boys among themselves can be arranged in 6! ways = 720 ways ∴ Total number of ways this can be done = 720 × 720 = 518400 ways

4.

In a class consisting of 5 girls and 6 boys, they are to be seated on 11 chairs in a row in such a way that all the boys always sit together. In how many different ways can this be done?1). 4320002). 4800003). 5280004). 518400

Answer»

Consider the 6 BOYS as one single unit

The total number of people will be 5 (girls) + 1 (All boys considered as one unit) = 6

These 6 people can be ARRANGED in 6! WAYS = 720 ways

The 6 boys among themselves can be arranged in 6! ways = 720 ways

∴ Total number of ways this can be done = 720 × 720 = 518400 ways

Discussion

5.	1). 1/32). 2/43). 1/64). 1
Answer» We know that, The prime NUMBERS occurring before 6 are 2 and 3 The SAMPLE SPACE (S) of the experiment = S = {1, 2, 3, 4, 5, 6} The Event (E) that only prime numbers show up on die = E = {2, 3, 5} Probability P(E) = n(e)/n(s) = 3/6 = 1/2 The probability of getting a prime number on a die is 1/2 or 2/4.

5.

1). 1/32). 2/43). 1/64). 1

Answer»

We know that,

The prime NUMBERS occurring before 6 are 2 and 3

The SAMPLE SPACE (S) of the experiment = S = {1, 2, 3, 4, 5, 6}

The Event (E) that only prime numbers show up on die = E = {2, 3, 5}

Probability P(E) = n(e)/n(s) = 3/6 = 1/2

The probability of getting a prime number on a die is 1/2 or 2/4.

Discussion

6.	Five letters are sent to different persons and addresses on the five envelopes are written ।at random. The probability that all the letters do not reach the correct destiny is1). 44/1202). 1/1203). 1/54). Cannot be determined
Answer» ? Probability = [No of Favorable OUTCOMES/No. of Total Outcomes] ⇒ No. of Total Outcomes = Total No. of ways in which 5 envelopes can be sent to 5 persons = 5! = 120 No. of Favorable Outcomes = when NONE of the letters reaches correct destiny Unfavorable CASES = when all the five envelope reaches to their destiny + when one of them reaches to its correct destiny + TWO of them reaches to its correct destiny + three of them reaches their correct place ⁵C₁ × [4! - ⁴C₁ × {3! – (³C₁ × 1 + 1)} + ⁴C₂ × 1 + 1] + ⁵C₂ × {3! – (³C₁ × 1 + 1)} + ⁵C₃ × 1 + 1 ? No. of Favorable Outcomes = Total Outcomes - Unfavorable Outcomes = 120 - [5 × 9 + 10 × 2 + 10 + 1] = 120 - 45 - 20 - 11 = 44 ⇒ Probability = 44/120

6.

Five letters are sent to different persons and addresses on the five envelopes are written ।at random. The probability that all the letters do not reach the correct destiny is1). 44/1202). 1/1203). 1/54). Cannot be determined

Answer»

? Probability = [No of Favorable OUTCOMES/No. of Total Outcomes]

⇒ No. of Total Outcomes = Total No. of ways in which 5 envelopes can be sent to 5 persons = 5! = 120

No. of Favorable Outcomes = when NONE of the letters reaches correct destiny

Unfavorable CASES = when all the five envelope reaches to their destiny + when one of them reaches to its correct destiny + TWO of them reaches to its correct destiny + three of them reaches their correct place

⁵C₁ × [4! - ⁴C₁ × {3! – (³C₁ × 1 + 1)} + ⁴C₂ × 1 + 1] + ⁵C₂ × {3! – (³C₁ × 1 + 1)} + ⁵C₃ × 1 + 1

? No. of Favorable Outcomes = Total Outcomes - Unfavorable Outcomes

= 120 - [5 × 9 + 10 × 2 + 10 + 1]

= 120 - 45 - 20 - 11 = 44

⇒ Probability = 44/120

Discussion

7.	Find the probability of getting at least 1 prime no. when 2 dices are rolled simultaneously?1). 1/22). 1/33). 3/44). 2/3
Answer» The sample space S for the above question is S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Prime nos. from 1 to 6 : 2, 3, 5 LET E be the EVENT: “at least one prime no. ROLLED in 2 dices” Let K be the event: “no prime no. is rolled” K = { (1,1), (1,4), (1,6), (4,1), (4,4), (4,6), (6,1), (6,4), (6,6) } P (E) = P (S) – P (K) ∴ P (E) = 1 – (9/36) = 1 – (1/4) = 3/4

7.

Find the probability of getting at least 1 prime no. when 2 dices are rolled simultaneously?1). 1/22). 1/33). 3/44). 2/3

Answer»

The sample space S for the above question is

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Prime nos. from 1 to 6 : 2, 3, 5

LET E be the EVENT: “at least one prime no. ROLLED in 2 dices”

Let K be the event: “no prime no. is rolled”

K = { (1,1), (1,4), (1,6), (4,1), (4,4), (4,6), (6,1), (6,4), (6,6) }

P (E) = P (S) – P (K)

∴ P (E) = 1 – (9/36) = 1 – (1/4) = 3/4

Discussion

8.	1). 503/5902). 497/5903). 500/6904). 400/690
Answer» TOTAL number of balls = 40 + 20 = 60 Total number of events = ⁶⁰C₂ Probability of getting both red balls = ⁴⁰C₂/⁶⁰C₂ Probability of getting UNDAMAGED ball = ⁵⁰C₂/⁶⁰C₂ Probability of getting undamaged red ball = ³²C₂/⁶⁰C₂ ∴ Probability of REQUIRE condition = [⁴⁰C₂ + ⁵⁰C₂– ³²C₂]/⁶⁰C₂ = [40 × 39 + 50 × 49 – 32 × 31]/[60 × 59] = [1560 + 2450 - 992]/3540 = 3018/3540 = 503/590

8.

1). 503/5902). 497/5903). 500/6904). 400/690

Answer»

TOTAL number of balls = 40 + 20 = 60

Total number of events = ⁶⁰C₂

Probability of getting both red balls = ⁴⁰C₂/⁶⁰C₂

Probability of getting UNDAMAGED ball = ⁵⁰C₂/⁶⁰C₂

Probability of getting undamaged red ball = ³²C₂/⁶⁰C₂

∴ Probability of REQUIRE condition = [⁴⁰C₂ + ⁵⁰C₂– ³²C₂]/⁶⁰C₂ = [40 × 39 + 50 × 49 – 32 × 31]/[60 × 59] = [1560 + 2450 - 992]/3540 = 3018/3540 = 503/590

Discussion

9.	A bag contains 5 red balls, 6 blue balls, 4 green balls and 3 black balls. If a ball is drawn at random find the probability that it is not blue1). 1/32). 2/33). 3/44). 3/5
Answer» PROBABILITY of an event = Number of FAVORABLE outcomes/Total number of outcomes Total no. of balls in the BAG = 5 + 6 + 4 + 3 = 18 The probability that the ball drawn is not blue = 1 – Probability of the ball being blue Probability of the ball being blue = ⁶C₁/¹⁸C₁ = 6/18 = 1/3 ⇒ Probability that the ball drawn is not blue = 1 – 1/3 = 2/3

9.

A bag contains 5 red balls, 6 blue balls, 4 green balls and 3 black balls. If a ball is drawn at random find the probability that it is not blue1). 1/32). 2/33). 3/44). 3/5

Answer»

PROBABILITY of an event = Number of FAVORABLE outcomes/Total number of outcomes

Total no. of balls in the BAG = 5 + 6 + 4 + 3 = 18

The probability that the ball drawn is not blue = 1 – Probability of the ball being blue

Probability of the ball being blue = ⁶C₁/¹⁸C₁ = 6/18 = 1/3

⇒ Probability that the ball drawn is not blue = 1 – 1/3 = 2/3

Discussion

10.	If three dice are thrown simultaneously, then the probability of getting a score of 5 is1). 5/2162). 1/63). 1/364). 1/72
Answer» Total number of possible outcomes when three DICE are thrown simultaneously = 6 × 6 × 6 = 216 Three NUMBERS add up to 5 ⇒ numbers can constitute the following sets: {1, 1, 3} or {1, 2, 2} [No die can show more than 3 and LESS than 1 for the three numbers to add up to 5] {1, 1, 3} can be rearranged in 3 WAYS as (1, 1, 3), (1, 3, 1) and (3, 1, 1) {1, 2, 2} can be rearranged in 3 ways as (1, 2, 2), (2, 1, 2) and (2, 2, 1) So, total number of ways in which three numbers on dice add up to 5 is 6. ∴ The REQUIRED probability = 6/216 = 1/36

10.

If three dice are thrown simultaneously, then the probability of getting a score of 5 is1). 5/2162). 1/63). 1/364). 1/72

Answer»

Total number of possible outcomes when three DICE are thrown simultaneously = 6 × 6 × 6 = 216

Three NUMBERS add up to 5 ⇒ numbers can constitute the following sets: {1, 1, 3} or {1, 2, 2}

[No die can show more than 3 and LESS than 1 for the three numbers to add up to 5]

{1, 1, 3} can be rearranged in 3 WAYS as (1, 1, 3), (1, 3, 1) and (3, 1, 1)

{1, 2, 2} can be rearranged in 3 ways as (1, 2, 2), (2, 1, 2) and (2, 2, 1)

So, total number of ways in which three numbers on dice add up to 5 is 6.

∴ The REQUIRED probability = 6/216 = 1/36

Discussion

11.	1). 3/42). 1/43). 1/24). 1
Answer» We know that, The Sample Space S = {5 + 5 + 5(HHH), 5 + 10 + 5(HTH), 5 + 5 + 10(HHT), 5 + 10 + 10(HTT), 10 + 5 + 5(THH), 10 + 10 + 10(TTT), 10 + 10 + 5(TTH), 10 + 5 + 10(THT)} = {15, 20, 20, 25, 20, 30, 25, 25} The event E of sum being less than or equal to 20 = {15, 20, 20, 20} Probability P(E) = n(e)/n(s) = 4/8 = 1/2 ∴ the probability of the sum being less than or equal to 20 is 1/2.

11.

1). 3/42). 1/43). 1/24). 1

Answer»

We know that,

The Sample Space S = {5 + 5 + 5(HHH), 5 + 10 + 5(HTH), 5 + 5 + 10(HHT), 5 + 10 + 10(HTT), 10 + 5 + 5(THH), 10 + 10 + 10(TTT), 10 + 10 + 5(TTH), 10 + 5 + 10(THT)} = {15, 20, 20, 25, 20, 30, 25, 25}

The event E of sum being less than or equal to 20 = {15, 20, 20, 20}

Probability P(E) = n(e)/n(s) = 4/8 = 1/2

∴ the probability of the sum being less than or equal to 20 is 1/2.

Discussion

12.	1). 1/82). 1/153). 1/64). 1/12
Answer» Probability of being the TRUTH that is said by the PERSON = 3/5 Probability of getting a FIVE in the dice = 1/6 ∴ The probability that it is actually a five = 3/5 × 1/6 = 1/10

12.

1). 1/82). 1/153). 1/64). 1/12

Answer»

Probability of being the TRUTH that is said by the PERSON = 3/5

Probability of getting a FIVE in the dice = 1/6

∴ The probability that it is actually a five = 3/5 × 1/6 = 1/10

Discussion

13.	1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A = Quantity B4). Quantity A ≤ Quantity B
Answer» QUANTITY A: PROBABILITY of 3 picked cards are from same suit = 4 × 3 cards picked from same suit ÷ 3 cards picked from deck Probability of 3 picked cards are from same suit = 4 × ¹³C₃ ÷ ⁵²C₃ = 4 × 286 ÷ 22100 ≈ 0.052 Quantity B: TOTAL no. of outcomes when 3 DICE rolled together = 6 × 6 × 6 = 216 Total no. of possibilities that all of them facing same no. = 6 Required Probability = 6/216 = 1/36 = 0.027 ∴ Quantity A > Quantity B

13.

1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A = Quantity B4). Quantity A ≤ Quantity B

Answer»

QUANTITY A:

PROBABILITY of 3 picked cards are from same suit = 4 × 3 cards picked from same suit ÷ 3 cards picked from deck

Probability of 3 picked cards are from same suit = 4 × ¹³C₃ ÷ ⁵²C₃ = 4 × 286 ÷ 22100 ≈ 0.052

Quantity B:

TOTAL no. of outcomes when 3 DICE rolled together = 6 × 6 × 6 = 216

Total no. of possibilities that all of them facing same no. = 6

Required Probability = 6/216 = 1/36 = 0.027

∴ Quantity A > Quantity B

Discussion

14.	Three unbiased coin are tossed. What is the probability of getting all heads?1). 5/92). 3/83). 1/84). 2/9
Answer» All the possible outcomes are HHH, HHT, HTH, THH, TTT, TTH, THT and HTT Total possible outcomes = 8 Number of times all head OBTAINED = 1 ∴ Probability of GETTING all head = 1/8

14.

Three unbiased coin are tossed. What is the probability of getting all heads?1). 5/92). 3/83). 1/84). 2/9

Answer»

All the possible outcomes are HHH, HHT, HTH, THH, TTT, TTH, THT and HTT

Total possible outcomes = 8

Number of times all head OBTAINED = 1

∴ Probability of GETTING all head = 1/8

Discussion

15.	A bag contains X red balls and 5 green balls. 2 balls are picked up from the bag at random, one after the other, without replacement, the probability of both balls being red is 3/7. What will be the value of X?1). 102). 153). 134). 20
Answer» Number of GREEN BALLS = 5 Number of Red balls = X ⇒ TOTAL number of balls = 5 + X Probability of both balls being red = 3/7 ⇒ ^XC₂/ ^{5 + X}C₂ = 3/7 Only 10 satisfies the value of X ∴ Number of red balls, X = 10

15.

A bag contains X red balls and 5 green balls. 2 balls are picked up from the bag at random, one after the other, without replacement, the probability of both balls being red is 3/7. What will be the value of X?1). 102). 153). 134). 20

Answer»

Number of GREEN BALLS = 5

Number of Red balls = X

⇒ TOTAL number of balls = 5 + X

Probability of both balls being red = 3/7

⇒ ^XC₂/ ^{5 + X}C₂ = 3/7

Only 10 satisfies the value of X

∴ Number of red balls, X = 10

Discussion

16.	A box contains red, green and blue colored balls. When one ball is drawn randomly from the box, the probability that it is either green or blue is 2/3. Also, the probability that the ball drawn randomly from the box is either red or green is 3/5. When two balls are drawn randomly from the box, the probability that both balls are of green color is 1/15. Find the total number of balls in the box.1). 152). 203). 304). 45
Answer» LET the NUMBER of RED, GREEN and blue balls be R, G and B respectively. When one ball is drawn randomly from the box, the probability that it is either green or blue = 2/3 = 10/15 The probability that the ball drawn randomly from the box is either red or green = 3/5 = 6/15 Let G + B = 10x R + G = 6x R + G + B =15x By SOLVING above equations, we get R = 5x G = 4x B = 6x Probability of picking two green balls = (4x/15x) × ((4x – 1)/(15x – 1)) But, when 2 balls are drawn at random, the probability that they both are of green color = 1/15 ⇒ (4x/15x) × ((4x – 1)/(15x – 1)) = 1/15 ⇒ 4(4x – 1) = (15x – 1) ⇒ 16x – 4 = 15x – 1 ⇒ 16x – 15x = 4 – 1 ⇒ x = 3 ∴ Total no. of balls in box is 45

16.

A box contains red, green and blue colored balls. When one ball is drawn randomly from the box, the probability that it is either green or blue is 2/3. Also, the probability that the ball drawn randomly from the box is either red or green is 3/5. When two balls are drawn randomly from the box, the probability that both balls are of green color is 1/15. Find the total number of balls in the box.1). 152). 203). 304). 45

Answer»

LET the NUMBER of RED, GREEN and blue balls be R, G and B respectively.

When one ball is drawn randomly from the box, the probability that it is either green or blue = 2/3 = 10/15

The probability that the ball drawn randomly from the box is either red or green = 3/5 = 6/15

Let G + B = 10x

R + G = 6x

R + G + B =15x

By SOLVING above equations, we get

R = 5x

G = 4x

B = 6x

Probability of picking two green balls = (4x/15x) × ((4x – 1)/(15x – 1))

But, when 2 balls are drawn at random, the probability that they both are of green color = 1/15

⇒ (4x/15x) × ((4x – 1)/(15x – 1)) = 1/15

⇒ 4(4x – 1) = (15x – 1)

⇒ 16x – 4 = 15x – 1

⇒ 16x – 15x = 4 – 1

⇒ x = 3

∴ Total no. of balls in box is 45

Discussion

17.	A bag contains 6 blue and 5 yellow balls. 4 balls are drawn successively without replacement. What is the probability that the balls are alternately of different colours?1). 7/332). 5/333). 3/114). 1/9
Answer» FINAL ANSWER = (Starting with a blue ball) OR (Starting with a YELLOW ball) Final Answer = (⁶C₁/¹¹C₁ × ⁵C₁/¹⁰C₁ × ⁵C₁/⁹C₁ × ⁴C₁/⁸C₁) + (⁵C₁/¹¹C₁ × ⁶C₁/¹⁰C₁ × ⁴C₁/⁹C₁ × ⁵C₁/⁸C₁) ∴ Final answer = (6/11 × 5/10 × 5/9× 4/8) + (5/11 × 6/10 × 4/9 × 5/8) ∴ The REQUIRED probability is 5/33

17.

A bag contains 6 blue and 5 yellow balls. 4 balls are drawn successively without replacement. What is the probability that the balls are alternately of different colours?1). 7/332). 5/333). 3/114). 1/9

Answer»

FINAL ANSWER = (Starting with a blue ball) OR (Starting with a YELLOW ball)

Final Answer = (⁶C₁/¹¹C₁ × ⁵C₁/¹⁰C₁ × ⁵C₁/⁹C₁ × ⁴C₁/⁸C₁) + (⁵C₁/¹¹C₁ × ⁶C₁/¹⁰C₁ × ⁴C₁/⁹C₁ × ⁵C₁/⁸C₁)

∴ Final answer = (6/11 × 5/10 × 5/9× 4/8) + (5/11 × 6/10 × 4/9 × 5/8)

∴ The REQUIRED probability is 5/33

Discussion

18.	All the face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn from it. What is the probability that the drawn card is an ace?1). 1/52). 1/43). 1/104). 2/10
Answer» All the face CARDS are removed. ∴ NUMBER of remaining cards = 52 - 12 = 40 Number of ace cards = 4 ∴ Required PROBABILITY = $(\frac{4}{{40}} = \frac{1}{{10}})$

18.

All the face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn from it. What is the probability that the drawn card is an ace?1). 1/52). 1/43). 1/104). 2/10

Answer»

All the face CARDS are removed.

∴ NUMBER of remaining cards = 52 - 12 = 40

Number of ace cards = 4

∴ Required PROBABILITY = $(\frac{4}{{40}} = \frac{1}{{10}})$

Discussion

Explore topic-wise InterviewSolutions in .

Rohan flips a coin, rolls a dice and draws a card. What is the probability of getting a head, 4 on dice and a card of king together?1). Chances are more than 1%2). Chances are less than 0.10%3). Chances are more than 2%4). Chances are more than 0.50%

Three packs of cards are thoroughly mixed and shuffled and 2 cards are drawn at random, one after the other. What is the probability that all of them are jacks?1). 43/23052). 4/33553). 7/13354). 98/1695

In a class consisting of 5 girls and 6 boys, they are to be seated on 11 chairs in a row in such a way that all the boys always sit together. In how many different ways can this be done?1). 4320002). 4800003). 5280004). 518400

1). 1/32). 2/43). 1/64). 1

Five letters are sent to different persons and addresses on the five envelopes are written ।at random. The probability that all the letters do not reach the correct destiny is1). 44/1202). 1/1203). 1/54). Cannot be determined

Find the probability of getting at least 1 prime no. when 2 dices are rolled simultaneously?1). 1/22). 1/33). 3/44). 2/3

1). 503/5902). 497/5903). 500/6904). 400/690

A bag contains 5 red balls, 6 blue balls, 4 green balls and 3 black balls. If a ball is drawn at random find the probability that it is not blue1). 1/32). 2/33). 3/44). 3/5

If three dice are thrown simultaneously, then the probability of getting a score of 5 is1). 5/2162). 1/63). 1/364). 1/72

1). 3/42). 1/43). 1/24). 1

1). 1/82). 1/153). 1/64). 1/12

1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A = Quantity B4). Quantity A ≤ Quantity B

Three unbiased coin are tossed. What is the probability of getting all heads?1). 5/92). 3/83). 1/84). 2/9

A bag contains X red balls and 5 green balls. 2 balls are picked up from the bag at random, one after the other, without replacement, the probability of both balls being red is 3/7. What will be the value of X?1). 102). 153). 134). 20

A bag contains 6 blue and 5 yellow balls. 4 balls are drawn successively without replacement. What is the probability that the balls are alternately of different colours?1). 7/332). 5/333). 3/114). 1/9

All the face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn from it. What is the probability that the drawn card is an ace?1). 1/52). 1/43). 1/104). 2/10