SBI PO 2018 Daily Quiz 03 Quantitative Aptitude “Probability”
SBI PO 2018 Quantitative aptitude Practice test. Daily Quiz for SBI 2018 online practice set. Daily tests with questions and answers for SBI exam. Topic tests for SBI online PT & Mains. English, Quantitative Aptitude , reasoning, General Knowledge solved practice questions sets for SBI Exam
1. A bag contains 8 apple and 6 orange. Four
fruits are drawn out one by one and not
replaced. What is the probability that they
are alternatively of different fruits?
A) 10/143
B) 15/120
C) 20/143
D) 26/110
E) None
2. In an interview the probability of Praveen
to got selected is 0.4. The probability of
Geetha to got selected is 0.5. The
probability of Sam to got selected is 0.6.
The probability of Suresh to got selected is
0.8. What is the probability that at least 2 of
them got selected on that day?
A) 0.806
B) 0.632
C) 0.688
D) 0.732
E) None
3. A basket contains 10 red ball and 15 white
ball. out of which 3 red and 4 white balls
are damaged. If two balls selected at
random, what is the probability that either
both are white balls or both are not
damaged?
A) 203/435
B) 313/300
C) 317/400
D) 203/300
E) None
4. A box contains tickets numbered from 1 to
16. 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 4?
A) 19/240
B) 11/240
C) 43/250
D) 9/80
E) None
5. Chance that Sheela tells truth is 35% and
for Ramesh is 75%. In what percent they
likely to contradict each other in the same
question?
A) 9/40
B) 15/25
C) 25/40
D) 23/40
E) None
6. Two dice are thrown simultaneously. What
is the probability of getting the sum of the
numbers as even?
A) 1/3
B) 2/3
C) 1/2
D) 3/4
E) None
7. A basket contains 8 Red and 6 Pink toys.
There is another basket which contains 7
Red and 8 Pink toys. One toy is to drawn
from either of the two baskets. What is the
probability of drawing a Pink toys?
A) 101/210
B) 85/156
C) 75/210
D) 120/156
E) None
8. Four persons are chosen at random from a
group of 3 men, 5 women and 4 children.
What is the probability of exactly two of
them being men?
A) 10/60
B) 12/55
C) 25/60
D) 13/60
E) None
9. A box contains 3 ballons of 1 shape, 4
ballons of 1 shape and 5 ballons of 1 shape.
Three ballons of them are drawn at random,
what is the probability that all the three are
of different shape?
A) 3/44
B) 5/22
C) 3/11
D) 10/22
E) None
10. 12 persons are seated around a round
table.What is the probability that two
particular persons sit together?
A) 2/11
B) 4/21
C) 8/21
D) 6/21
E) None
Answer
1. Option C
Solution:
Fruits can be drawn in two format
AOAO and OAOA
Apple drawn 1st P=8/14*6/13*7/12*5/11
Orange drawn 1st P=6/14*8/13*5/12*7/11
Adding both we get
2[8*7*6*5/14*13*12*11]=2*(10/143)=20/
143.
2. Option A
Solution:
Required probability=1 – no one got
selected – 1 got selected
No one got selected = (1-0.4) x (1-0.5) x
(1-0.6) x (1-0.8) = 0.024
1 got selected= 0.4 x ((1-0.5) x (1-0.6) x (1-
0.8) ) + 0.5 x ((1-0.4) x (1-0.6) x (1-0.8))
+0.6 x ((1-0.4) x (1-0.5) x (1-0.8)) + 0.8 x
((1-0.4) x (1-0.5) x (1-0.6))
= 0.016 + 0.024 + 0.036 + 0.096 = 0.17
So, Required probability = 1 – 0.024 – 0.17= 0.806
3. Option B
Solution: |
P(A) = 15c2 / 25c2, P(B) = 18c2 / 25c2
P(A∩B) = 11c2 / 25c2
P(A∪B) = P(A) + P(B) – P(A∩B) => (15c2
/ 25c2)+( 18c2 / 25c2)-( 11c2 /
25c2)=406/600==>203/300
4. Option A
Solution:
From 1 to 16, there are 4 numbers which
are multiple of 4
1st 2 are multiple of 4, and one any other
number from (16-4) = 12 tickets
4c2*12c1/16c3 = 72/560
2nd all are multiples of 4.
4c3/16c3=4/560
Add both 72/560+4/560.=76/560
5. Option D
Solution:
P(A) = 35/100=7/20 and P(B) =
75/100=3/4.
Now they are contradicting means one lies
and other speaks truth. So,
Probability = 7/20*1/4 + 13/20 * 3/4
=7/80+39/80=46/80=23/40
6. Option C
Solution:
Throw two dice n(s)=36
E is nos sum is even.
Hence E={( 1,1 ),(1,3 ) (1,5 ) (2,2 ) , (2,4 ) ,
(2,6 ), …………( 6,2 ), (6,4 ), ( 6,6 )}
n(E)= 18
Thus required probability= 18/36= 1/2
7. Option A
Solution:
Probability of one basket =1/2
1st Basket Pink toy probability =1/2*
(6c1/14c1)
2nd Basket Pink toy probability = 1/2*
(8c1/15c1)
Adding both (1/2*6/14) + (1/2*4/15)
3/14+4/15=101/210
8. Option B
Solution:
Total People = 3 + 5 + 4 = 12
n(s)=12c4
Probability of exactly two men and two
from others
N(e) = 3c2*9c2
⇒ P= (3c2*9c2)/12c4=>12/55
9. Option C
Solution:
Total=3+4+5=12
n(s)=12c3=220
n(e)=3c1 * 4c1 * 5c1 =60
p=60/220=3/11
10. Option A
Solution:
In a circle of n different persons, the total
number of arrangements possible = (n – 1)!
n(S) = (12 – 1) = 11 !
Taking two persons as a unit, total persons
= 11
Therefore no. of ways for these 11 persons
to around the circular table = (11 – 1)! =
10!
In any unit, 2 particular person can sit in 2!
ways.
Hence total number of ways that any three
person can sit,
=n(E) = 10! * 2!
Therefore P (E) = probability of three
persons sitting together = n(E) / n(S)
= (10! * 2!)/11! = 2/11
