1.What is the probability of getting at least one 3 in 2 throws of a dice?
1. 1/36
2. 5/36
3. 11/36
4. 35/36
Answer & Explanation
Sol : Option 3
Total possible outcomes = 36.
Favorable cases = 11 i.e. (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6). (1, 3) (2, 3) (4, 3) ,(5, 3) ,(6, 3)
Therefore, Reqd. Probability = 11 / 36
2.The probability of A’s winning a game of chess against B is 2/3. What is the probability that A will win at least 1 of a total of two games?
1. 1/27
2. 19/27
3. 2/27
4. 8/9
Answer & Explanation
Sol : Option 4
Reqd. Probability = 1 – (A not winning even one Game out of 3) =1 - (1/3)2 = 1 - (1/9) = (8/9)
3.What is the probability that a non-leap year will have 53 Mondays?
1. 52/53
2. 51/52
3. 1
4. None of these
Answer & Explanation
Sol : Option 4
Non Leap Year = 52 Weeks and one Day.
Prob. of one days Day being Monday = 1 / 7
Therefore, Reqd. Prob. = 1 / 7
Directions for questions 4 to 6: Read the following information and answer the questions that follow. The probability that A will pass the examination is 1/2 and the probability that B will pass the examination is 1/3.
4.What is the probability that both A and B will pass the examination?
1. 1/6
2. 1
3. 2/3
4. 1/3
Answer & Explanation
Sol : Option 1
Probability that A will pass in exam = 1/2
So, Probability that A will fail in exam = 1/2.
Probability that B will pass in exam = 1/3
So, Probability that B will fail in exam = 2/3.
Probability that both will pass in the exam = Probability that A will pass and probability that B will pass
= 1/2 x 1/3 = 1/6
5.What is the probability that only 1 person [either A or B] will pass the examination?
1. 1
2. 1/2
3. 1/3
4. 2/3
Answer & Explanation
Sol : Option 2
Probability that only one person will pass is when A passes and B fails or A fails and B passes.
= (1/2) x (2/3) + (1/2) x (1/3) = 1/2
6.What is the probability that at least one person will pass the examination?
1. 1
2. 1/2
3. 1/3
4. 2/3
Answer & Explanation
Sol : Option 4
Probability that at least one person, will pass, so the possibilities
can be (A pass and B fails), or (A fails and B pass) or (Both A and B pass)
(1/2) x (2/3) + (1/2) x (1/3) + (1/2) x (1/3) = 2/3
7.Three cards are drawn together from a pack of 52 cards at random. What is the probability that all the cards are Diamonds?
1. 4C3 / 2C3
2. 13C3 / 52C3
3. 26C3 / 52C3
4. 8C3 / 52C3
Answer & Explanation
Sol : Option 2
There are 13 diamonds. Three diamonds out of 13 diamonds can be taken out in 13C3 ways.
Total number of sample spaces = 52C3
Required probability = 13C3 / 52C3
8.A bag contains 8 blue balls and 6 black balls. Three balls are drawn one by one with replacement. What is the probability that all the 3 balls are black?
1. 27 / 343
2. 1 / 343
3. 18 / 343
4. None of these
Answer & Explanation
Sol : Option 1
P (of black ball in first attempt) = 6 / 14 = 3 / 7
Here probability will remain same for the next two attempt.
Therefore, Required Probability = (3/7) x (3/7) x (3/7) = 27 / 343
9.One bag contains 8 blue balls and 6 Green balls; another bag contains 7 blue balls and 5 green balls. If one ball is drawn from each bag, determine the probability that both are blue?
1. 1/2
2. 1/3
3. 1/4
4. 1/5
Answer & Explanation
Sol : Option 2
Bag I Bag II
8 blue 7 blue
6 green 5 green Probability of getting blue ball from bag I = 8 / 14
Probability of getting blue ball from bag II = 7 / 12
Hence reqd. Prob. = (8/14) x (7/12) = (1/3)
10.1 ball is drawn at random from a box containing 4 red balls, 5 white balls and 6 blue balls, what is the probability that the ball is a red ball?
1. 1/7
2. 2/15
3. 4/15
4. 1/15
Answer & Explanation
Sol : Option 3
Total possible outcomes = 15. (i.e. One out of 15 balls).
Favorable outcomes (one out of 4 red balls) = 4.
Reqd. Probability = 4 / 15
11. The probability that a number selected at random from the first 50 natural numbers is a composite number is -.
A. 21/25
B. 17/25
C. 4/25
D. 8/25
E. 9/25
Answer & Explanation
Answer: Option B
Explanation:
The number of exhaustive events = ⁵⁰C₁ = 50.
We have 15 primes from 1 to 50.
Number of favourable cases are 34.
Required probability = 34/50 = 17/25.
12. A coin is tossed five times. What is the probability that there is at the least one tail?
A. 31/32
B. 1/16
C. 1/2
D. 1/32
E. None of these
Answer & Explanation
Answer: Option A
Explanation:
Let P(T) be the probability of getting least one tail when the coin is tossed five times.
= There is not even a single tail.
i.e. all the outcomes are heads.
= 1/32 ; P(T) = 1 - 1/32 = 31/32
13. If a number is chosen at random from the set {1, 2, 3, ...., 100}, then the probability that the chosen number is a perfect cube is -.
A. 1/25
B. 1/2
C. 4/13
D. 1/10
E. 9/13
Answer & Explanation
Answer: Option A
Explanation:
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25.
14. Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is -.
A. 1/2
B. 16/19
C. 4/5
D. 17/20
E. 3/5
Answer & Explanation
Answer: Option D
Explanation:
n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(Eá´œP) = 10/20 + 8/20 - 1/20 = 17/20
15. If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is -.
A. 1/4
B. 1/2
C. 3/4
D. 3/5
E. None of these
Answer & Explanation
Answer: Option B
Explanation:
The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
16. If four coins are tossed, the probability of getting two heads and two tails is -.
A. 3/8
B. 6/11
C. 2/5
D. 4/5
E. 5/11
Answer & Explanation
Answer: Option A
Explanation:
Since four coins are tossed, sample space = 24
Getting two heads and two tails can happen in six ways.
n(E) = six ways
p(E) = 6/2^4 = 3/8
17. If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is -.
A. 19/52
B. 17/52
C. 5/13
D. 4/13
E. 9/26
Answer & Explanation
Answer: Option D
Explanation:
P(Sá´œK) = P(S) + P(K) - P(S∩K), where S denotes spade and K denotes king.
P(Sá´œK) = 13/52 + 4/52 - 1/52 = 4/13
18. If six persons sit in a row, then the probability that three particular persons are always together is -.
A. 1/20
B. 3/10
C. 1/5
D. 4/5
E. 2/5
Answer & Explanation
Answer: Option C
Explanation:
Six persons can be arranged in a row in 6! ways.
Treat the three persons to sit together as one unit
then there four persons and they can be arranged in 4! ways.
Again three persons can be arranged among them selves in 3! ways.
Favourable outcomes = 3!4! Required probability = 3!4!/6! = 1/5
19. A bag contains 7 green and 8 white balls. If two balls are drawn simultaneously, the probability that both are of the same colour is -.
A. 8/15
B. 2/5
C. 3/5
D. 11/15
E. 7/15
Answer & Explanation
Answer: Option E
Explanation:
Drawing two balls of same color from seven green balls can be done in ⁷C₂ ways.
Similarly from eight white balls two can be drawn in ⁸C₂ ways.
P = ⁷C₂/¹⁵C₂ + ⁸C₂/¹⁵C₂ = 7/15
20. A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is -.
A. 25/42
B. 9/20
C. 10/23
D. 41/42
E. 1/42
Answer & Explanation
Answer: Option A
Explanation:
Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.
21.A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.
A) 1/2 B) 7/15
C) 8/15 D) 1/9
Answer & ExplanationAnswer: B) 7/15
Explanation:
Let S be the sample space
Then n(S) = no of ways of drawing 2 balls out of (6+4) =10C2 10 =10*92*1 =45
Let E = event of getting both balls of same colour
Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)
=6C2+4C2 = 6*52*1+4*32*1= 15+6 = 21
Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15
22.Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A) 1/2 B) 3/5
C) 9/20 D) 8/15
Answer & ExplanationAnswer: C) 9/20
Explanation:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/n(S) = 9/20.
23.Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?
A) 52/221 B) 55/190
C) 55/221 D) 19/221
Answer & ExplanationAnswer: C) 55/221
Explanation:
We have n(s) =52C2 52 = 52*51/2*1= 1326.
Let A = event of getting both black cards
B = event of getting both queens
A∩B = event of getting queen of black cards
n(A) = 52*51/2*1 = 26C2 = 325, n(B)= 26*25/2*1= 4*3/2*1= 6 and n(A∩B) = 4C2 = 1
P(A) = n(A)/n(S) = 325/1326;
P(B) = n(B)/n(S) = 6/1326 and
P(A∩B) = n(A∩B)/n(S) = 1/1326
P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221
24.A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
A) 2/91 B) 1/22
C) 3/22 D) 2/77
Answer & ExplanationAnswer: A) 2/91
Explanation:
Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3 =15*14*13/3*2*1= 455.
Let E = event of getting all the 3 red balls.
n(E) = 5C3 = 5*4/2*1 = 10.
=> P(E) = n(E)/n(S) = 10/455 = 2/91.
25.Two dice are tossed. The probability that the total score is a prime number is:
A) 5/12 B) 1/6
C) 1/2 D) 7/9
Answer & ExplanationAnswer: A) 5/12
Explanation:
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = n(E)/n(S) = 15/36 = 5/12.
26.In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
A) 2/7 B) 5/7
C) 1/5 D) 1/2
Answer & ExplanationAnswer: A) 2/7
Explanation:
Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
P(E)=n(E)/n(S)=10/35=2/7
27.Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
A) 3/20 B) 29/34
C) 47/100 D) 13/102
Answer & ExplanationAnswer: D) 13/102
Explanation:
Let S be the sample space.
Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326.
Let E = event of getting 1 spade and 1 heart.
n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169.
P(E) = n(E)/n(S) = 169/1326 = 13/102.
28.One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
A) 3/13 B) 1/13
C) 3/52 D) 9/52
Answer & ExplanationAnswer: A) 3/13
Explanation:
Clearly, there are 52 cards, out of which there are 12 face cards.
P (getting a face card) = 12/52=3/13.
29Three unbiased coins are tossed.What is the probability of getting at least 2 heads?
A) 1/4 B) 1/2
C) 3/4 D) 1/3
Answer & ExplanationAnswer: B) 1/2
Explanation:
Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.
P(E) = n(E) / n(S)= 4/8= 1/2
30.In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
A) 21/46 B) 1/5
C) 3/25 D) 1/50
Answer & ExplanationAnswer: A) 21/46
Explanation:
Let , S - sample space
E - event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= 25C3 = 2300.
n(E) = 10C1×15C2 = 1050.
P(E) = n(E)/n(s) = 1050/2300 = 21/46
31.Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.
A) 7/18 B) 14/35
C) 8/18 D) 7/35
Answer & ExplanationAnswer: A) 7/18
Explanation:
Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18
32.A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?
A) 136/345 B) 17/87
C) 316/435 D) 158/435
Answer & ExplanationAnswer: C) 316/435
Explanation:
n(s)=C230
Let A be the event of getting two oranges and
B be the event of getting two non-defective fruits.
and (A∩B) be the event of getting two non-defective oranges
P(A)=20C2/30C2, P(B)=22C2/30C2 and P(A∩B)=15C2/30C2
P(A∪B)=P(A)+P(B)−P(A∩B)
= 20C2/30C2+22C2/30C2−15C2/30C2=316/435
33.Three unbiased coins are tossed. What is the probability of getting at most two heads?
A) 3/4 B) 7/8
C) 1/2 D) 1/4
Answer & ExplanationAnswer: B) 7/8
Explanation:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
34.What is the probability of getting 53 Mondays in a leap year?
A) 1/7 B) 3/7
C) 2/7 D) 1
Answer & ExplanationAnswer: C) 2/7
Explanation:
1 year = 365 days . A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364days
366 – 364 = 2 days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 days = 2/7
35.What is the probability of getting a sum 9 from two throws of a dice?
A) 1/2 B) 3/4
C) 1/9 D) 2/9
Answer & ExplanationAnswer: C) 1/9
Explanation:
In two throws of a die, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
P(E) =n(E)/n(S)=4/36=1/9.
36.A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident
A) 30/100 B) 35/100
C) 45/100 D) 50/100
Answer & ExplanationAnswer: B) 35/100
Explanation:
Let A = Event that A speaks the truth
B = Event that B speaks the truth
Then P(A) = 75/100 = 3/4
P(B) = 80/100 = 4/5
P(A-lie) = 1−34= 1/4
P(B-lie) = 1−45= 1/5
Now, A and B contradict each other =[A lies and B true] or [B true and B lies]
= P(A).P(B-lie) + P(A-lie).P(B)
= (3/5*1/5)+(1/4*4/5)=720
= (7/20*100)= 35%
37.A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.
A) 551/15134 B) 1/2
C) 552/15379 D) 1/9
Answer & ExplanationAnswer: A) 551/15134
Explanation:
Total number of elementary events = 50C5
Given,third ticket =30
=> first and second should come from tickets numbered 1 to 29 = 29C2 ways and remaining two in 20C2 ways.
Therfore,favourable number of events = 29C2*20C2
Hence,required probability = 29C2*20C2/50C5 =551 / 15134
38.From well shuffled standard pack of 52 playing cards,one card is drawn. What is the probability that it is either a red card or black card?
A) 1/2 B) 3/4
C) 1 D) 1/3
Answer & ExplanationAnswer: C) 1
Explanation:
P(red cards)=26/52
P(black cards)=26/52
P(red or black cards)=26/52+26/52=1
39.Find the probability that one ball is white when two balls are drawn at random from a basket that contains 9 red, 7 white and 4 black balls.
A) 18/95 B) 18/190
C) 1/2 D) 91/190
Answer & ExplanationAnswer: D) 91/190
Explanation:
Total number of elementary events = 20C2 ways =190
There are 7 white balls out of which one white can be drawn in 7C1 ways and one ball from remaining 13 balls can be drawn in 13C1 ways.
So,favourable events = 7C1*13C1
Therfore,required probability = (7C1*13C1)/20C2=91/90
40.In a single throw with 2 dices, what is probability of neither getting an even number on one and nor a multiple of 3 on other?
A) 11/36 B) 25/36
C) 5/6 D) 1/6
Answer & ExplanationAnswer: B) 25/36
Explanation:
We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and
E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4)
n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36
41.A bag contains 1100 tickets numbered 1, 2, 3, … 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?
a. 291/1100
b. 292/1100
c. 290/1100
d. 301/1100
Answer: c
Explanation:
Numbers which dont have 2 from 1 to 9 = 8
Numbers which don’t have 2 from 10 to 99:
Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with 9 ways. Total 72 numbers.
Numbers which don’t have 2 from 100 to 999 =_ _ _ = 8 × 9 × 9 = 648
Numbers which don’t have 2 from 1000 to 1099 =10_ _ = 9 × 9 = 81
Finally 1100 does not have 2. So 1.
Total number with no 2 in them = 8 + 72 + 648 + 81 + 1= 810
Tickets with 2 in them = 1100 – 810 = 290
Required probability = 290 / 1100
42. In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.
a. 1565
b. 2256
c. 2456
d. 1243
Answer: b
Explanation:
Maximum 3 men can be played which means there can be 0, 1, 2, 3 men in the team.
(5C0×11C11)+(5C1×11C10)(5C0×11C11)+(5C1×11C10) + (5C2×11C9)+(5C3×11C8)(5C2×11C9)+(5C3×11C8) = 2256
43.Four people each roll a four die once. Find the probability that at least two people will roll the same number ?
a. 5/18
b. 13/18
c. None of the given choices
d. 1295/1296
Answer: b
Explanation:
The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 x 5 x 4 x3
Now total possibilities of rolling a dice = 6^4
The probability that a no one gets the same number =
(6*5*4*3)/6^4 = 5/18
So the probability that at least two people gets same number =1-5/18 = 13/18
44.In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.
a. 1/3
b. 1/2
c. 5/9
d. 17/36
Answer: Their sum can be 3,4,6,8,9,12
Explanation: For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 – n) ways.
Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.
So probability is (20/36)=(5/9)
45.A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?
a.24/71
b.22/91
c.24/91
d.22/71
Answer:c
Explaination:This is a case of without replacement. We have to multiply two probabilities. 1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.
6C1/14C1 × 8C1/1C1 =24/91.
46.There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red?
a.312/16807
b.314/16907
c.322/13904
d.331/19807
Answer:
Explanation:At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.
case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21
Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
= 312/16807
47.2/3rd of the balls in a bag are blue, the rest are pink. if 5/9th of the blue balls and 7/8th of the pink balls are defective, find the total number of balls in the bag given that the number of non defective balls is 146.
a.431
b.24
c.221
d.23
Answer:c
Explanation: let the total balls be x 2/3x=blue & 1/3x=pink non defective balls=146 total no. of balls=defective+non defective hence defective=total-non defective(146) (2/3x)*5/9+(1/3x)*7/8=t-146 0.66203x=t-146 146=0.3303x x=432
48.From a bag containing 8 green and 5 red balls, three are drawn one after the other .the probability of all three balls beings green if the balls drawn are replaced before the next ball pick and the balls drawn are not replaced, are respectively?
a.512/2197, 336/2197
b.512/2197, 336/1716
c.336/2197, 512/2197
d.336/1716, 512/1716
Answer:a
Explanation:
The Probabilities of getting with replacement=8/13*8/13*8/13=512/2197 The Probabilities of getting without replacement =8/13*7/12*6/11=336/2197
49.There are two bags, one of which contains 5 red and 7 white balls and the other 3 red and 12 white balls. A ball is to be drawn from one or other of the two bags ; find the chances of drawing a red ball.
a.55/102
b.17/21
c.37/120
d.7/8
Answer:c
Explanation:
P(Getting a red ball from the first bag): (1/2)*(5/12) P(Getting a red ball from the 2nd bag): (1/2)*(3/15) P(Drawing a red ball): (1/2)*(5/12)+(1/2)*(3/15)= 37/120 (Ans).
50.3 dice are rolled. What is the probability that you will get the sum of the no’s as 10?
a.27/216
b.25/216
c.10/216
d.1/8
Answer: d
Explanation:
total events = 6*6*6=216 possible cases for sum equal to 10 are (1,3,6)-6 combinations (1,4,5)-6 combinations (2,3,5)-6 combinations (2,4,4)-3 combinations (3,3,4)-3 combinations (2,2,6)-3 combinations so total combinations are 27 so probability will be 27/216 = 1/8
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