When two coins are tossed simultaneously, what are the chances of getting at least one tail?
A bag contains 3 red balls and 8 blacks ball and another bag contains 5 red balls and 7 blacks balls, one ball is drawn at random from either of the bag, find the probability that the ball is red.
Probability = probability of selecting the bag and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264
12 persons are seated at a circular table. Find the probability that 3 particular persons always seated together.
total probability = (121)! = 11!
Desired probability = (10 – 1)! = 9!
So, p = (9! *3!) /11! = 3/55
P and Q are two friends standing in a circular arrangement with 10 more people. Find the probability that exactly 3 persons are seated between P and Q.
Fix P at one point then number of places where B can be seated is 11.
Now, exactly three persons can be seated between P and Q, so only two places where Q can be seated. So, p = 2/11
A basket contains 5 black and 8 yellow balls. Four balls are drawn at random and not replaced. What is the probability that they are of different colours alternatively.
sol⇒ BYBY + YBYB = (5/13)*(8/12)*(4/11)*(7/10) + (8/13)*(5/12)*(7/11)*(4/10)
= 56/429
Direction(Q6 – Q8):
A bag contains 6 red balls and 8 green balls. Two balls are drawn at random one after one with replacement. What is the probability that
Q.
Both the balls are green
P = (8/14)*(8/14)
A bag contains 6 red balls and 8 green balls. Two balls are drawn at random one after one with replacement. What is the probability that
Q.
First one is green and second one is red
P = (8/14)*(6/14)
A bag contains 6 red balls and 8 green balls. Two balls are drawn at random one after one with replacement. What is the probability that
Q.
Both the balls are red
P = (6/14)*(6/14)
Find the probability that in a leap year, the numbers of Mondays are 53?
In a leap year there are 52 complete weeks i.e. 364 days and 2 more days. These 2 days can be SM, MT, TW, WT, TF, FS, and SS.
So P = 2/7
A urn contains 4 red balls, 5 green balls and 6 white balls, if one ball is drawn at random, find the probability that it is neither red nor white.
5c1/15c1 = 1/3
A box contains tickets numbered from 1 to 24. 3 tickets are to be chosen to give 3 prizes. What is the probability that at least 2 tickets contain a number which is multiple of 3?
From 1 to 24, there are 8 numbers which are multiple of 3
Case 1: 2 are multiple of 3, and one any other number from (248) = 16 tickets
^{8}C_{2} * ^{16}C_{1} / ^{24}C_{3} = 56/253
Case 2: all are multiples of 3.
^{8}C_{3} / ^{24}C_{3} = 7/253
Add both cases.
A box contains 6 blue, 5 green and 4 red balls. Two balls are drawn at random. What is the probability that there is no red ball?
Total balls = 15
Not red means green or blue i.e. any of (5+6) = 11 balls
So prob. = ^{11}C_{2} / ^{15}C_{2}
From a pack of 52 cards, 2 cards are drawn at random. What is the probability that both cards are black card or heart card?
Prob. of black card:
^{26}C_{2} / ^{52}C_{2} = 25/102
Prob. of heart card:
^{13}C_{2} / ^{52}C_{2} = 3/51
Add both cases.
Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black or both are jacks?
There are total 26 cards black, and 4 jacks in which 2 are black jacks
So case 1: both are black
^{26}C_{2} / ^{52}C_{2}
case 2: both are jack
^{4}C_{2} / ^{52}C_{2}
Add both cases.
But now 2 black jacks have been added in both cases, so subtracting their prob. :
^{2}C_{2} / ^{52}C_{2}
So 325/1326 + 6/1326 – 1/1326
From a group of 3 men, 4 women and 2 children, 4 people are to be chosen to form a committee. What is the probability that the committee contains 1 each of men, women and children?
Case 1: Prob. when 2 men, 1 woman and 1 child
^{3}C_{2} * ^{4}C_{1} * ^{2}C_{1} / ^{9}C_{4} = 4/21
Case 2: Prob. when 1 man, 2 women and 1 child
^{3}C_{1} * ^{4}C_{1} * ^{2}C_{1} / ^{9}C_{4} = 2/7
Case 3: Prob. when 1 man, 1 woman and 2 children
^{3}C_{1} * ^{4}C_{1} * ^{2}C_{2} / ^{9}C_{4} = 2/21
A box contains 25 bulbs out of which 5 are defective. 3 bulbs are to be delivered to a customer. What is the probability that he get one defective bulb?
There are 4 red balls, 5 white and 3 green balls in a basket. 3 balls are chosen at random. What is the probability that there is at most 1 green ball?
Case 1: 0 green ball means all three red or white balls
^{9}C_{3} / ^{12}C_{3}
Case 2: 1 green ball and two red or white balls
^{9}C_{2} * ^{3}C_{1} / ^{12}C_{3} Add both cases.
A bag contains 3 red, 4 green and 3 yellow balls. If 2 balls are drawn at random, what is the probability that they are of different color?
This will be = 1 prob.(both are same in color)
Prob. of both same in color = [^{3}C_{2} + ^{4}C_{2} + ^{3}C_{2}]/ ^{10}C_{2} = 12/45
So required prob. = 1 – 12/45
There are 4 black balls and 6 white balls. 2 balls are drawn one by one without replacement. What is the probability that the balls are same in color?
When both black, prob. = 4/10 * 3/9
When both white, prob. = 6/10 * 5/9 Add both cases.
A bag contains 5 red balls and 4 green balls. What is the probability that both balls are same in color?
Case 1: both red
^{5}C_{2} / ^{9}C_{2}
Case 2: both green
^{4}C_{2} / ^{9}C_{2}
Add both cases
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