Online exam in Probability For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLA
Probability-Test 4
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Subject :- Quantitative Aptitude
Chapter :- Probability-Test 4
Questions :- 25
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A box contains 30 electric bulbs, out of which 8 are defective. Four bulbs are chosen at random from this box. Find the probability that at least one of them is defective?
Correct
1 – 22c4/30c4 = 1 – 209/783 = 574/783
Incorrect
1 – 22c4/30c4 = 1 – 209/783 = 574/783
Question 2 of 21
2. Question
1 points
Two person A and B appear in an interview. The probability of A’s selection is
1/5 and the probability of B’s selection is 2/7. What is the probability that only one of them is selected?
Correct
A selects and B rejects + B selects and A rejects
= (1/5)*(5/7) + (4/5)*(2/7) = 13/35
Incorrect
A selects and B rejects + B selects and A rejects
= (1/5)*(5/7) + (4/5)*(2/7) = 13/35
Question 3 of 21
3. Question
1 points
A 4- digit number is formed by the digits 0, 1, 2, 5 and 8 without repetition. Find the
probability that the number is divisible by 5.
Correct
Total possibility = 5*4*3*2
Favourable outcomes = 2*4*3*2 (to be
divisible by 5 unit digit can be filled with only 0
or 5, so only two possibilities are there, then the
remaining can be filled in 4, 3 and 2 ways respectively)
so probability = 2/5
Incorrect
Total possibility = 5*4*3*2
Favourable outcomes = 2*4*3*2 (to be
divisible by 5 unit digit can be filled with only 0
or 5, so only two possibilities are there, then the
remaining can be filled in 4, 3 and 2 ways respectively)
so probability = 2/5
Question 4 of 21
4. Question
1 points
A bag contains 6 red balls and 8 green balls. 2 balls are drawn at random one by one
with replacement. Find the probability that both the balls are green
Correct
(8c1)/(14c1) * (8c1)*(14c1) = 16/49
Incorrect
(8c1)/(14c1) * (8c1)*(14c1) = 16/49
Question 5 of 21
5. Question
1 points
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.
Correct
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
Incorrect
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
Question 6 of 21
6. Question
1 points
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?
Correct
(6/13)*(3/8) + (7/13)*(5/8) = 53/104
Incorrect
(6/13)*(3/8) + (7/13)*(5/8) = 53/104
Question 7 of 21
7. Question
1 points
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.
Correct
Second year students = 200
so, P = 200/700 = 2/7
Incorrect
Second year students = 200
so, P = 200/700 = 2/7
Question 8 of 21
8. Question
1 points
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?
Correct
1 – [13/52 + 4/52 – 1/52] = 9/13
Incorrect
1 – [13/52 + 4/52 – 1/52] = 9/13
Question 9 of 21
9. Question
1 points
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?
Correct
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
Incorrect
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
Question 10 of 21
10. Question
1 points
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?
Correct
(4c1 + 4c1 + 4c1)/(52c3)
Incorrect
(4c1 + 4c1 + 4c1)/(52c3)
Question 11 of 21
11. Question
1 points
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.
Correct
Total outcomes = (10 -1)! = 9!
Favourable outcomes = (9 -1)!*2!
So p = 2/9
Incorrect
Total outcomes = (10 -1)! = 9!
Favourable outcomes = (9 -1)!*2!
So p = 2/9
Question 12 of 21
12. Question
1 points
Find the probability that in a random arrangement of the letter of words in the
word ‘PROBABILITY’ the two I’s come together.
Correct
Total outcomes = 11!/(2!*2!)
favourable outcomes = (10!*2!)/(2!*2!)
p = 2/11
Incorrect
Total outcomes = 11!/(2!*2!)
favourable outcomes = (10!*2!)/(2!*2!)
p = 2/11
Question 13 of 21
13. Question
1 points
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.
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.
Correct
Probability = probability of selecting the bag
and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264
Incorrect
Probability = probability of selecting the bag
and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264
Question 15 of 21
15. Question
1 points
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.
Correct
(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
Incorrect
(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
Question 16 of 21
16. Question
1 points
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.
Correct
Probability = probability of selecting the bag and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264
Incorrect
Probability = probability of selecting the bag and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264
Question 17 of 21
17. Question
1 points
12 persons are seated at a circular table. Find the probability that 3 particular persons
always seated together.
Correct
total probability = (12-1)! = 11!
Desired probability = (10 – 1)! = 9!
So, p = (9! *3!) /11! = 3/55
Incorrect
total probability = (12-1)! = 11!
Desired probability = (10 – 1)! = 9!
So, p = (9! *3!) /11! = 3/55
Question 18 of 21
18. Question
1 points
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.
Correct
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
Incorrect
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
Question 19 of 21
19. Question
1 points
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.