Probability Questions for IBPS PO
Probability Questions for IBPS PO 2016 Exam
1. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5. (IBPS PO Mains 2016)
(a) 3/5
(b) 1/5
(b) 2/5
(d) 4/5
(e) None of these
2. A box contains 5 green, 4 yellow and 3 white marbles. Three marbles are drawn at random. What is the probability that they are not of the same colour ? (IBPS PO Mains 2016)
(a) 40/44
(b) 44/41
(c) 41/44
(d) 40/39
(e) 44/39
3. From a pack of 52 cards, 3 cards are drawn together at random, What is the probability of both the cards are king ? (IBPS PO Mains 2016)
\(\left( a \right) \quad \frac { 1 }{ 5225 } \)
\(\left( b \right) \quad \frac { 1 }{ 5525 } \)
(c) 5225
\(\left( d \right) \quad \frac { 1 }{ 525 } \)
(e) None of these
Probability Questions for IBPS PO 2016 Answers
1. (b)
5 digit number = 5! = 120
Divisible by 5 then the last digit should be 0
Then the remaining position have the possibility = 4! = 24
P = (4!/5!) = 24/120 = 1/5
2. (c)
n(E) = 5C3 + 4C3 + 3C3 =10 + 4 +1 = 15
n(S) = 12C3= 220
\(\mathrm { P } = \frac { \mathrm { n } ( \mathrm { E } ) } { \mathrm { n } ( \mathrm { S } ) } = \frac { 15 } { 220 } = \frac { 3 } { 44 }\)
Required probability
\(= 1 – \frac { 3 } { 44 } = \frac { 41 } { 44 }\)
3. (b)
\(\mathrm { n } ( \mathrm { S } ) = ^ { 52 } \mathrm { C } _ { 3 } = \frac { 132600 } { 6 } = 22100\)
\(\mathrm { n } ( \mathrm { E } ) = ^ { 4 } \mathrm { C } _ { 3 } = \frac { 24 } { 6 } = 4\)
\(\mathrm { p } = \frac { 4 } { 22100 } = \frac { 1 } { 5525 }\)
Probability Questions for IBPS PO 2015 Exam
Study the following information carefully to answer the questions that follow-
A committee of five members is to be formed out of 5 Males, 6 Females and 3 Children. In how many different ways can it be 58. done if-? (IBPS PO Prelim 2015)
1. The committee should consist of 2 Males, 2 Females and 1 Child?
(a) 450
(b) 225
(c) 55
(d) 90
(e) None of these
2. The committee should include all the 3 Childs?
(a) 90
(b) 180
(c) 21
(d) 55
(e) None of these
3. In how many different ways can 4 boys and 3 girls be arranged in a row such that all the boys stand together and all the girls stand together ? (IBPS PO Prelim 2015)
(a) 75
(b) 576
(c) 288
(d) 24
(e) None of these
4. In how many different ways can the letters of the word ‘PRIDE’ be arranged ? (IBPS PO Main 2015)
(a) 60
(b) 120
(c) 15
(d) 360
(e) None of these
5. There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls?(IBPS PO Main 2015)
\(\left( a \right) \quad \frac { 191 }{ 1547 } \)
\(\left( b \right) \quad \frac { 180 }{ 1547 } \)
\(\left( c \right) \quad \frac { 280 }{ 1547 } \)
\(\left( d \right) \quad \frac { 189 }{ 1547 } \)
Probability Questions for IBPS PO 2015 Answers
1. (a)
Number of ways = 5C2 × 6C2 × 3C1= 450
2. (d)
Number ofways = 11C2 × 5C2 = 55.
3. (c)
Required number of ways = 4! × 31 × 2! = 288.
4. (b)
‘PRIDE’ has five different letters.
So, it can be arranged in 5 ! = 120 ways
5. (c)
Total possible outcomes = 17C5
\(= \frac { 17 \times 16 \times 15 \times 14 \times 13 } { 1 \times 2 \times 3 \times 4 \times 5 } = 6188\)
Total favourable outcomes = 8C2 × 4C1 × 5C2
\(= \frac { 8 \times 7 } { 1 \times 2 } \times 4 \times \frac { 5 \times 4 } { 1 \times 2 } = 28 \times 4 \times 10 = 1120\)
Required probability
\(= \frac { 1120 } { 6188 } = \frac { 280 } { 1547 }\)
Probability Questions for IBPS PO 2014 Exam
1. In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together? (IBPS PO/MT 2014)
(a) 360
(b) 480
(c) 720
(d) 5040
(e) None of these
2. 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: (IBPS PO/MT 2014)
\(\left( a \right) \quad \frac { 21 }{ 46 } \)
\(\left( b \right) \quad \frac { 25 }{ 117 } \)
\(\left( c \right) \quad \frac { 1 }{ 50 } \)
\(\left( d \right) \quad \frac { 3 }{ 25 } \)
(e) None of these
3. Gauri went to the stationery and bought things worth ’25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items? (IBPS PO/MT 2014)
(a) ₹ 15
(b) ₹ 15.70
(c) ₹ 19.70
(d) ₹ 20
(e) None of these
Probability Questions for IBPS PO 2014 Answers
1. (c)
The word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
∴ Required number of ways = (120 × 6) = 720.
2. (a)
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= 25C3
\(= \frac { ( 25 \times 24 \times 23 ) } { ( 3 \times 2 \times 1 ) } = 2300\)
n(E) = (10C1 × 15C2)
\(= \left[ 10 \times \frac { ( 15 \times 14 ) } { ( 2 \times 1 ) } \right]\)
= 1050
∴ \(\mathrm { P } ( \mathrm { E } ) = \frac { \mathrm { n } ( \mathrm { E } ) } { \mathrm { n } ( \mathrm { S } ) } = \frac { 1050 } { 2300 } = \frac { 21 } { 46 }\)
3. (c)
Let the amount taxable purchases be Rs. x.
Then, 6 % of
\(x = \frac { 30 } { 100 }\)
\(\Rightarrow x = \left( \frac { 30 } { 100 } \times \frac { 100 } { 6 } \right) = 5\)
∴ Cost of tax free items = Rs.[25 – (5 + 0.30)] = ₹ 19.70
Probability Questions for IBPS PO 2012 Exam
1. In how many different ways can the letters of the word ‘THERAPY’ be arranged so that the vowels never come together ? (IBPS PO/MT 2012)
(a) 720
(b) 1440
(c) 5040
(d) 3600
(e) 4800
2. A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour ? (IBPS PO/MT 2012)
\(\left( a \right) \quad \frac { 41 }{ 190 } \)
\(\left( b \right) \quad \frac { 21 }{ 190 } \)
\(\left( c \right) \quad \frac { 59 }{ 190 } \)
\(\left( d \right) \quad \frac { 99 }{ 190 } \)
\(\left( e \right) \quad \frac { 77 }{ 190 } \)
Probability Questions for IBPS PO 2012 Answers
1. (d)
No. of vowels in the word THERAPY
= 2 i.e. E and A
In such cases we treat the group of two vowels as one entity or one letter because they are supposed to always come together. Thus, the problem reduces to arranging 6 letters i.e. T, H, R, P, Y and \(\boxed { EA }\) in 6 vacant places.
No. of ways 6 letters can be arranged in 6 places = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
But the vowels can be arranged themselves in 2 different ways by interchanging their position. Hence, each of the above 720 arrangements can be written in 2 ways.
∴ Required no. of total arrangements when two vowels are together = 720 × 2 = 1440
Total no. of arrangements of THERAPY = 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
No. of arrangement when vowels do not come together
= 5040 – 1440 = 3600
2. (d)
No. of ways of getting 2 white balls = 13C2
No. of ways of getting 2 black balls = 7C2
Probability of getting 2 same colour ball
\(=\frac { Probability\quad of\quad 2white\quad balls\quad or\quad Probability\quad of\quad 2\quad Black\quad balls }{ Total\quad number\quad of\quad balls\quad drawn } \)
Probability Questions for IBPS PO 2011 Exam
Study the given information carefully to answer the questions that follow.
An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles.
1. If four marbles are drawn at random, what is the probability that two are blue and two are red?
\(\left( a \right) \quad \frac { 10 }{ 1001 } \)
\(\left( b \right) \quad \frac { 9 }{ 14 } \)
\(\left( c \right) \quad \frac { 17 }{ 364 } \)
\(\left( d \right) \quad \frac { 2 }{ 7 } \)
(e) None of these
2. If eight marbles are drawn at random, what is the probability that there are equal number of marbles of each colour ?
\(\left( a \right) \quad \frac { 4 }{ 7 } \)
\(\left( b \right) \quad \frac { 361 }{ 728 } \)
\(\left( c \right) \quad \frac { 60 }{ 1001 } \)
\(\left( d \right) \quad \frac { 1 }{1 } \)
(e) None of these
3. If two marbles are drawn at random, what is the probability that both are red or at least one is red ?
\(\left( a \right) \quad \frac { 26 }{ 91 } \)
\(\left( b \right) \quad \frac { 1 }{ 7 } \)
\(\left( c \right) \quad \frac { 199 }{ 364 } \)
\(\left( d \right) \quad \frac { 133 }{ 191 } \)
(e) None of these
4. If three marbles are drawn at random, what is the probability that at least one is yellow ?
\(\left( a \right) \quad \frac { 1 }{ 3 } \)
\(\left( b \right) \quad \frac { 199 }{ 364 } \)
\(\left( c \right) \quad \frac { 165 }{ 364 } \)
\(\left( d \right) \quad \frac { 3 }{ 11 } \)
(e) None of these
5. If three marbles are drawn at random, what is the probability that none is green ? (IBPS PO 2011)
\(\left( a \right) \quad \frac { 2 }{ 7 } \)
\(\left( b \right) \quad \frac { 253 }{ 728 } \)
\(\left( c \right) \quad \frac { 10 }{ 21 } \)
\(\left( d \right) \quad \frac { 14 }{ 91 } \)
(e) None of these
Probability Questions for IBPS PO 2011 Answers
1. (a)
According to question.
\(n ( S ) = ^ { 14 } C _ { 4 } = \frac { 14 ! } { ( 14 – 4 ) ! 4 ! } = \frac { 14 ! } { 10 ! 4 ! }\)
\(=\frac { 14\times 13\times 12\times 11 }{ 4\times 3\times 2\times 1 } =1001\)[∵ \(^{ { n } }C_{ { r } }=\frac { n! }{ (n-r)!r! } \)]
and
\(n ( E ) = ^ { 5 } C _ { 2 } \times ^ { 2 } C _ { 2 } = \frac { 5 ! } { ( 5 – 2 ) ! 2 ! } \times \frac { 2 ! } { ( 2 – 2 ) ! 2 ! }\)
\(= \frac { 5 \times 4 } { 2 \times 1 } \times \frac { 2 \times 1 } { 1 \times 2 \times 1 } = 10\)
∴ Required probability
\( = \frac { n ( E ) } { n ( S ) } = \frac { 10 } { 1001 }\)
2. (c)
According to the question
\(n ( S ) = ^ { 14 } C _ { 8 } = \frac { 14 ! } { ( 14 – 8 ) ! 8 ! } \times \frac { 14 ! } { 6 ! 8 ! }\)
\(= \frac { 14 \times 13 \times 12 \times 11 \times 10 \times 9 } { 6 \times 5 \times 4 \times 3 \times 2 \times 1 } = 3003\)
and n(E) = 4C2 × 5C2 × 2C2 × 3C2
\(= \frac { 4 ! } { ( 4 – 2 ) ! 2 ! } \times \frac { 5 ! } { ( 5 – 2 ) ! 2 ! } \times \frac { 2 ! } { ( 2 – 2 ) ! 2 ! } \times \frac { 3 ! } { ( 3 – 2 ) ! 2 ! }\)
\(= \frac { 4 ! } { 2 ! 2 ! } \times \frac { 5 ! } { 3 ! 2 ! } \times \frac { 2 ! } { 0 ! 2 ! } \times \frac { 3 ! } { 1 ! 2 ! }\)
\(= \frac { 4 \times 3 } { 2 \times 1 } \times \frac { 5 \times 4 } { 2 \times 1 } \times \frac { 1 } { 1 } \times \frac { 3 } { 1 } = 180\)
∴ Required probability
\(= \frac { n ( E ) } { n ( S ) } = \frac { 180 } { 3003 } = \frac { 60 } { 1001 }\)
3. (e)
According to the question
\(n ( S ) = ^ { 14 } C _ { 2 } = \frac { 14 ! } { ( 14 – 2 ) ! 2 ! } = \frac { 14 \times 13 } { 2 \times 1 } = 91\)
∴ Probability of at least one red ball
\(= 1 – \frac { 12 C _ { 2 } } { 14 C _ { 2 } } = 1 – \frac { 66 } { 91 } = \frac { 91 – 66 } { 91 } = \frac { 25 } { 91 }\)
4. (b)
According to the question
\(n ( S ) = ^ { 14 } C _ { 3 } = \frac { 14 ! } { ( 14 – 3 ) ! 3 ! } = \frac { 14 \times 13 \times 12 } { 3 \times 2 \times 1 } = 364\)
∴ Required probability
\(= 1 – \frac { 11 } { 14 } \mathrm { C } _ { 3 } = 1 – \frac { 165 } { 364 } = \frac { 364 – 165 } { 364 } = \frac { 199 } { 364 }\)
5. (e)
According to the question
\(n ( S ) = ^ { 14 } C _ { 3 } = \frac { 14 ! } { ( 14 – 3 ) ! 3 ! } = \frac { 14 \times 13 \times 12 } { 3 \times 2 \times 1 } = 364\)
and
\(n ( E ) = ^ { 10 } C _ { 3 } = \frac { 10 ! } { ( 10 – 3 ) ! 3 ! } = \frac { 10 \times 9 \times 8 } { 3 \times 2 \times 1 } = 120\)
∴ Required probability
\(= \frac { n ( E ) } { n ( S ) } = \frac { 120 } { 364 } = \frac { 30 } { 91 }\)
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