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A bag contains 5 red balls and 7 blue balls. Two balls are drawn at random without replacement, and then find the probability of that one is red and other is blue.
Explanation :
(First red ball is drawn and then blue ball is drawn) + (first blue ball is drawn and then red ball is drawn)
(5/12)*(7/11) + (7/12)*(5/11) = 35/66
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 = (12-1)! = 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(6 – 8): 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-
Both the balls are green.
P = (8/14)*(8/14)
First one is green and second one is red.
P = (8/14)*(6/14)
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 six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6. Find the probability that the number is divisible by 4.
For a number to be divisible by 4, the last two digit should be divisible by 4.
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 +24 +24 +24 + 24+ 24+24+24 = 192
So p = 192/720 = 4/15
A bag contains 6 red balls and 7 white balls. Another bag contains 5 red balls and 3 white balls. One ball is selected from each. Find the probability that one ball is red and one is white?
(6/13)*(3/8) + (7/13)*(5/8) = 53/104
A lottery is organised by the college ABC through which they will provide scholarship of rupees one lakhs to only one student. There are 100 fourth year students, 150 third year students, 200 second year students and 250 first year students. What is the probability that a second year student is choosen.
Second year students = 200
so, P = 200/700 = 2/7
A card is drawn from a pack of 52 cards. The card is drawn at random; find the probability that it is neither club nor queen?
1 – [13/52 + 4/52 – 1/52] = 9/13
A box contains 50 balls, numbered from 1 to 50. If three balls are drawn at random with replacement. What is the probability that sum of the numbers are odd?
There are 25 odd and 25 even numbers from 1 to 50.
Sum will be odd if = odd + odd + odd, odd + even + even, even + odd + even, even+ even + odd
P= (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2)
=4/8 = ½
From a pack of cards, if three cards are drawn at random one after the other with replacement, find the probability that one is ace, one is jack and one is queen?
(4c1 + 4c1 + 4c1)/(52c3)
A and B are two persons sitting in a circular arrangement with 8 other persons. Find the probability that both A and B sit together.
Total outcomes = (10 -1)! = 9!
Favourable outcomes = (9 -1)!*2!
So p = 2/9
Find the probability that in a random arrangement of the letter of words in the word ‘PROBABILITY’ the two I’s come together.
Total outcomes = 11!/(2!*2!)
favourable outcomes = (10!*2!)/(2!*2!)
p = 2/11
In a race of 12 cars, the probability that car A will win is 1/5 and of car B is 1/6 and that of car C is 1/3. Find the probability that only one of them won the race.
1/5 + 1/6 + 1/3= 7/10 (all events are mutually exclusive)
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