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probability questions and answers
Posted Date: 18 Jun 2008 Resource Type: Articles/Knowledge Sharing Category: Placement Papers
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Posted By: anil kamani Member Level: Gold
Rating:  Points: 3
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Probability Questions and Answers
Q1 Given that E and F are events such that P(E) = 0.6 and P(F) = 0.3 and P(E n F) = 0.2 find P(E/F) and P(F/E)
Solution: P(E/F) = P(E n F) = 0.2 = 2
P( F ) 0.3 3
P(F/E) = P(E n F) = 0.2 = 1
P( E ) 0.6 3 Ans.
Compute P(A/B), if P(B) = 0.5 and P(A n B) = 0.32
P(A/B) = P(A n B) = 0.32 = 16
P( B ) 0.5 25 Ans.
If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4 find
1) P(A n B) 2) P(A/B) 3) P (A U B)
P(B/A) = P(A n B) P(A/B) = P(A n B) P(A U B)= 0.8 + 0.5 – 0.32
P(A) P(B)
0.4 = P(A n B) = 0.32 = 0.98 Ans.
0.8 0. 5
0.32 = P(A n B) = 0.64
Q2) Evaluate P (A U B) if 2P(A)= P(B)= 5/13 and P(A/B)= 2/5
Solution:
2P(A) = 5/13
P(A/B) = 5/26
P(A/B) = P(A n B)
P(B)
2/5 = P(A n B)
5/13
P(A n B ) = 2/5 * 5/13 = 2/13
P(A U B) = P(A) + P(B) – P( A n B)
= 5/26+ 5/13 – 2/13
= 11/26 Ans.
Q3) If P(A)= 6/11, P(B)= 5/11 and P(A U B) = 7/11 find i) P(AnB)
ii) P(A/B) iii) P(B/A)
Solution:
i) P(A U B) = P (A) + P(B) – P (A n B)
7/11 = 6/11 + 5/11 – P(AnB)
P(AnB) = 4/11
ii) P(A/B) = P(AnB)
P(B)
= 4/11 = 4/5
5/11
iii) P(B/A) = P(AnB)
P(A)
= 4/11 = 2/3 Ans.
6/11
Q4) Find P(E/F) if a coin is tossed thrice
E: head on third toss F: heads on first 2 tosses
S = { HHH,HHT,HTH,HTT,TTT,THH,THT,TTH } = 8
P(F) = { HHH,HHT} = 2/8
P(E) = { HHH} = 1/8
P(E U F) = { HHH } = 1/8
P(E/F) = 1/8 = 1/2 Ans.
2/8
Q5) Find P(E/F) if a coin is tossed thrice
E: at least 2 heads F: at most 2 heads
P(F) = {HHT,HTH,HTT,TTT,THH,THT,TTH} = 7/8
P(E) = { HHH, HHT,HTH,THH} = 4/8
P(E U F) = { HHT,HTH,THH } = 3/8
P(E/F) = 3/8 = 3/7 Ans.
7/8
Q6) Find P(E/F) if a coin is tossed thrice
E: at most 2 tails F: at least one tail.
P(F) = {HHT,HTH,HTT,TTT,THH,THT,TTH } = 7/8
P(E) = {HHH,HHT,HTH,HTT,THH,THT,TTH} = 7/8
P(E U F) = {HHT,HTH,HTT,THH,THT,TTH } = 6/8
P(E/F) = 6/8 = 6/7 Ans.
7/8
Q7) Find P(E/F) if 2 coins are tossed once
E: tail appears on one coin F: one coin shows head
S = { HH,HT,TH,TT } = 4
P(F) = { TH,HT} = 2/4
P(E) = { HT,TH} = 2/4
P(E U F) = { HT, TH } = 2/4
P(E/F) = 2/4 = 1 Ans.
2/4
Q8) Find P(E/F) if a die is thrown 3 times
E: 4 appears on the 3rd toss F: 6 and 5 appears respectively
on first 2 tosses.
P(F) = { (6,5,1) (6,5,2) (6,5,3) (6,5,4) (6,5,5) (6,5,6) } = 6/216
P(E) = { 6,5,4 } = 1 / 216
P(E U F) = { 6,5,4 } = 1 / 216
P(E/F) = 1/216 = 1 / 6 Ans.
6/216
Q9) Find P(E/F) if a father, mother and son line up at random
for a family picture.
E: son on one end F: father in middle
S = { (FMS) (FSM) (MSF)(MFS)(SMF)(SFM) } = 6
P(F) = { ( MFS ) (SFM) } = 2 / 6
P(E) = { (MFS)(SFM) } = 2 / 6
P(E U F) = {(MFS)(SFM) } = 2 / 6
P(E/F) = 2 / 6 = 1 Ans.
2 / 6
Q10) A black and red die are rolled
Find the probability of obtaining a sum greater than 9, given
that the black die resulted in a 5
S = 36
P(F) = { (1,5)(2,5)(3,5)(4,5)(5,5)(6,5) } = 6 / 36
P(E) = { (5,5)(6,5) = 2 / 36
P(E U F) = {(5,5) (6,5) } = 2 / 36
P(E/F) = 2 / 36 = 1 / 3 Ans.
6 / 36
Q11) A black and red die are rolled
Find the probability of obtaining a sum of 8, given that
the red die resulted in a number less than 4
S = 36
P(F) = 18 / 36
P(E) = 2 / 36
P(E U F) = 2 / 36
P(E/F) = 2 / 36 = 1 / 9 Ans.
18 / 36
Q12) A fair die is rolled. Consider events E = { 1,3,5 }, F = { 2,3 } and
G = { 2,3,4,5 } find
i) P(E / F)
P (E) = 3 / 6
P(F) = 2 / 6
P(E U F) = 1 / 6
P(E/F) = 1 / 6 = 1 / 2 Ans.
2 / 6
ii) P( F/E )
P(F / E) = 1 / 6 = 1 / 3 Ans.
3 / 6
Q13) A fair die is rolled. Consider events E = { 1,3,5 }, F = { 2,3 } and
G = { 2,3,4,5 } find
i ) P(E / G)
P (E) = 3 / 12
P(G) = 4 / 12
P(E U G) = 2 / 12
P(E/G) = 2 / 12 = 1 / 2 Ans.
4 / 12
ii) P( G/E )
P(G / E) = 2 / 12 = 2 / 3 Ans.
3 / 12
Q14) A fair die is rolled. Consider events E = { 1,3,5 }, F = { 2,3 } and
G = { 2,3,4,5 } find
i ) P { (E U F) / G }
P (E U F) n G = { 1,2,3,5} n { 2,3,4,5 }
= 3/6
P (G ) = 4 / 6
P{ E U F) /G } = P ( E U F ) n G = 3 / 6 = 3 / 4 Ans.
P ( G ) 4 / 6
ii) P (E n F ) / G
P ( E n F ) n G = { 3 } n { 2,3,4,5 }
= 1 / 6
P ( G ) = 4 / 6
P{ E n F) /G } = P ( E n F ) n G = 1 / 6 = 1 / 4 Ans.
P ( G ) 4 / 6
Q15) Assuming that each born child is equally likely to be a boy or a girl.
If a family has 2 children, what is the probability that both are girls
given that i) the youngest is a girl ii) at least one is a girl
Solution:
i) P(E) : both are girls P(F): the youngest is a girl
S = { (G G ) ( G B ) (B G ) ( B B) }
P(F) = { ( GG )( BG ) } = 2 / 4
P(E) = { ( GG ) } = 1/ 4
P(E n F) = (GG) = 1 / 4
P ( E/F) = P(E n F) = 1 / 4 = 1 / 2 Ans.
P(F) 2 / 4
ii) P(E) : both are girls P(F): at least one is a girl
S = { (G G ) ( G B ) (B G ) ( B B) }
P(F) = { ( GG )( BG ) ( GB) } = 3 / 4
P(E) = { ( GG ) } = 1 / 4
P(E n F) = (GG) = 1 / 4
P ( E/F) = P(E n F) = 1 / 4 = 1 / 3 Ans.
P(F) 3 / 4
Q16) An instructor has a question bank consists of 300 easy True/ False
questions, 200 difficult True/False questions, 500 easy multiple choice
questions and 400 difficult multiple choice questions. If a question is
selected , what is the probability that it will be a easy question given
that it is a multiple choice question ?
Solution:
P(E) : easy question P(F): multiple choice question
S = 300 + 200 + 500 + 400 = 1400
P(F) = { 500 + 400 }= 900 / 1400
P(E) = { 300 + 500 } = 800 / 1400
P(E n F) = 500 / 1400
P ( E/F) = P(E n F) = 500 / 1400 = 5 / 9 Ans.
P(F) 900 / 1400
Q17) Given that the 2 numbers appearing on throwing 2 dice are different.
Find the probability of the event “ the sum of the numbers on the dice
is 4 “
Solution:
P(E): the sum is 4 P(F) : both the numbers are different
S = 36
P(F) = 30 / 36 { leaving the doublet }
P(E) = { (1,3)(3,1)(2,2) } = 3 / 36
P(E n F) = 2 / 36
P ( E/F) = P(E n F) = 2 / 36 = 1 / 15 Ans.
P(F) 30 / 36
Q18) Consider the experiment of throwing a die, if a multiple of 3 comes up
throw the die again and if any other number comes, toss a coin. Find
the probability of the event “ the coin shows the tail” given that
“at least one die shows a 3”
Solution:
P(E): the coin shows the tail P(F): at least one die shows a 3
S ={ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
(1,H)(2,H)(4,H)(5,H)(1,T)(2,T)(4,T)(5,T) } = 20
P(F) = 7 / 20
P(E) = 4 / 20
P(E n F) = 0
P ( E/F) = P(E n F) = 0 = 0 Ans.
P(F) 7 / 20
Q19) If P(A) =3/5 and P(B) = 1/5 find P(AnB) if A and B are
independent events.
Solution: As A and B are independent events
P(AnB) = P(A) . P(B)
= 3 * 1 = 3 Ans.
5 5 25
Q18) Let E and F be events with P(E) = 3/5, P(F) = 3/10 and
P(E n F) = 1/5. Are E and F independent ?
If A and B are independent events, then
P(EnFB) = P(E) . P(B)
1 = 3 * 3
5 5 10
1 ? 9
5 50 E and F are not independent.
Q20) 2 cards are drawn without replacement from a pack of 52 cards.
Find the Probability that both are black.
Solution:
P(E): 1st card is black P(F): 2nd card is also black
We have to find P(EnF)
P(EnF) = P(E) . P(F)
= 26 * 25 ( Without replacement)
52 51
= 25 Ans.
102
Q21) A box of oranges is inspected by selecting 3 oranges without
Replacement. If all the 3 oranges are good, the box is approved foe
Sale otherwise it is rejected. Find the probability that a box containing
15 Oranges out of which 12 are good and 3 are bad ones will be
approved for sale
Solution: 12 good ones + 3 bad ones
P(E): 1st one is good P(F): 2nd one is good P(G): 3rd one good
All the 3 events are independent
We have to find P(EnFnG)
P(EnFnG) = P(E) . P(F) . P(G)
= 12 * 11 * 10 (Without replacement)
15 14 13
= 44 Ans.
91
Q22) A fair coin and an unbiased die are tossed. Let A be the event
“head appears on the coin” and B be the event “3 on the die”.
Check weather A and B are independent events are not.
Solution:
S ={ (H,1)(H,2)(H,3)(H,4)(H,5)(H,6)(T,1)(T,2)(T,3)(T,4)(T,5)(T,6) } = 12
P(A): head on the coin P(B): 3 on the die
The 2 events are independent or not ?
We have to check
P(AnB) = P(A) . P(B)
1 = 6 * 2
12 12 12
1 = 1
12 12
Clearly A and B are Independent events.
Q23) A die marked 1,2,3 in red and 4,5,6 in green is tossed.
Let A be the Event “the number is even” and B be the event
“the number is red” are A and B Independent ?
Solution:
S = 1 , 2 , 3 , 4 , 5 , 6 = 6
P(A): the number is even P(B): the number is red
The 2 events are independent or not ?
We have to check
P(AnB) = P(A) . P(B)
1 = 3 * 3
6 6 6
1 ? 1
6 4
Clearly A and B are not Independent events.
Q24) Given that the events A and B are such that P(A) = 1 /2 and
P(B) = p and P(A U B) = 3/5. Find p if they are i) mutually exclusive
ii) Independent.
Solution: i) mutually exclusive
If A and B are mutually exclusive, then
P(AUB) = P(A) + P(B)
3 = 1 + p
5 2
p = 3 – 1 = 1 Ans.
5 2 10
ii) A and B are Independent events.
P(AUB) = P(A) + P(B) – P (A nB)
P(AUB) = P(A) + P(B) – P (A) . P(B)
3 = 1 + p - 1 * p
5 2 2
3 – 1 = 2p – p
5 2 p
p = 1 , p = 1/5 Ans.
2 10
Q25) Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4
Find i) P(A n B) ii) P(A U B) iii) P(A / B) iv) P(B/A)
Solution: A and B are Independent events.
i) P (A nB) = P (A) . P(B)
= 0.3 * 0.4 = 0.12
ii) P(AUB) = P(A) + P(B) – P (A nB)
= 0.3 + 0.4 - 0.12 = 0.58
iii) P(A/B) = P(AnB)
P(B)
= 0.12 = 0.03
0.4
iv) P(B/A) = P(AnB)
P(A)
= 0.12 = 0.04 Ans.
0.3
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