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Probability problems

Posted by Ravi Kumar at Tuesday, March 15, 2011
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Probability problems

Probability problems


Problems:

1)An biased die is tossed.Find the probability of getting a
multiple of 3?

Sol: Here we have sample space S={1,2,3,4,5,6}.
Let E be the event of getting a multiple of 3.
Then E={3,6}.
P(E) =n(E)/n(S).
n(E) =2,
n(S) =6.
P(E) =2/6
P(E) =1/3.

2)In a simultaneous throw of a pair of dice,find the
probability of getting a total more than 7?

Sol: Here we have sample space n(S) =6*6 =36.
Let E be the event of getting a total more than 7.
={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),
(5,3),(6,2),(4,5),(5,4),
(5,5),(4,6),(6,4)}
n(E) =15
P(E) = n(E)/n(S)
= 15/36.
P(E) = 5/12.

3)A bag contains 6 white and 4 black balls .Two balls are
drawn at random .Find the probability that they are of the
same colour?

Sol: Let S be the sample space.
Number of ways for drawing two balls out of 6 white and
4 red balls = 10C2
=10!/(8!*2!)
= 45.
n(S) =45.
Let E =event of getting both balls of the same colour.
Then
n(E) =number of ways of drawing ( 2balls out of 6) or
(2 balls out of 4).
= 6C2 +4C2
= 6!/(4!*2!) + 4!/(2! *2!)
= 6*5/2 +4 *3/2
=15+6 =21.
P(E) =n(E)/n(S) =21/45 =7/45.

4)Two dice are thrown together .What is the probability that
the sum of the number on
the two faces is divisible by 4 or 6?
Sol: n(S) = 6*6 =36.
E be the event for getting the sum of the number on the two
faces is divisible by 4 or 6.
E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)
(5,1)(5,3)(6,2)(6,6)}
n(E) =14.
Hence P(E) =n(E)/n(S)
= 14/36.
P(E) = 7/18

5)Two cards are drawn at random from a pack of 52 cards What
is the probability that either both are black or both are
queens?
Sol: total number of ways for choosing 2 cards from
52 cards is =52C2 =52 !/(50!*2!)
= 1326.
Let A= event of getting bothe black cards.
Let B= event of getting bothe queens
AnB=Event of getting queens of black cards
n(A) =26C2.
We have 26 black cards from that we have to choose 2 cards.
n(A) =26C2=26!/(24!*2!)
= 26*25/2=325
from 52 cards we have 4 queens.
n(B) = 4C2
= 4!/(2!* 2!) =6
n(AnB) =2C2. =1
P(A) = n(A) /n(S) =325/1326
P(B) = n(B)/n(S) = 6/1326
P(A n B) = n(A n B)/n(S) = 1/1326
P(A u B) = P(A) +P(B) -P(AnB)
= 325/1326 + 6/1326 -1/1326
= 330/1326
P(AuB) = 55/221

6)Two diced are tossed the probability that the total score
is a prime number?

Number of total ways n(S) =6 * 6 =36
E =event that the sum is a prime number.
Then E={(1,1)(1,2)(1,4)(1,6)(2,1)(2,3)(2,5)(3,2)(3,4)(4,1)
(4,3)(5,2)(5,6)(6,1)(6,5)}
n(E) =15
P(E) =n(E)/n(S)
= 15/36
P(E) = 5/12


7)Two dice are thrown simultaneously .what is the probability
of getting two numbers whose product is even?

Sol : In a simultaneous throw of two dice ,we have n(S) = 6*6
= 36
E=Event of getting two numbers whose product is even
E={(1,2)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)
(3,4)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,4)(5,6)(6,1)
(6,2)(6,3)(6,4)(6,5)(6,6)}
n(E) = 27
P(E) = n(E)/n(S)
= 27 /36
P(E) =3/4
probability of getting two numbers whose product is even is
equals to 3/4.

8)In a lottery ,there are 10 prozes and 25 blanks.A lottery is
drawn at random. what is the probability of getting a prize ?

Sol: By drawing lottery at random ,we have n(S) =10C1+25C1
= 10+25
= 35.
E =event of getting a prize.
n(E) =10C1 =10
out of 10 prozes we have to get into one prize .The number of
ways 10C1.
n(E) =10
n(S) =35
P(E) =n(E)/n(S)
=10/35
= 2/7
Probability is 2/7.

9)In a class ,30 % of the students offered English,20 % offered
Hindi and 10 %offered Both.If a student is offered at random,
what is the probability that he has offered English or Hindi?

Sol:English offered students =30 %.
Hindi offered students =20%
Both offered students =10 %
Then only english offered students E =30 -10
=20 %
only Hindi offered students S =20 -10 %
= 10 %
All the students =100% =E +S +E or S
100 =20 +10 + E or S +E and S
Hindi or English offered students =100 -20-10-10
=60 %
Probability that he has offered English or Hindi =60/100= 2/5

10) A box contains 20 electricbulbs ,out of which 4 are defective,
two bulbs are chosen at random from this box.What is the
probability that at least one of these is defective ?

Sol: out of 20 bulbs ,4 bulbs are defective.
16 bulbs are favourable bulbs.
E = event for getting no bulb is defective.
n(E) =16 C 2
out of 16 bulbs we have to choose 2 bulbs randomly .so the number
of ways =16 C 2
n(E) =16 C2
n(S) =20 C 2
P(E) =16 C2/20C2
= 12/19
probability of at least one is defective + probability of one
is non defective =1
P(E) + P(E) =1
12/19 +P(E) =1
P(E’) =7/19

11)A box contains 10 block and 10 white balls.What is the
probability of drawing two balls of the same colour?

Sol: Total number of balls =10 +10
=20 balls
Let S be the sample space.
n(S) =number of ways drawing 2 balls out of 20
= 20 C2
= 20 !/(18! *2!)
= 190.
Let E =event of drawing 2 balls of the same colour.
n(E) =10C2+ 10C2
= 2(10 C2)
= 90
P(E) =n(E)/n(S)
P(E) =90/190
= 9/19

12) A bag contains 4 white balls ,5 red and 6 blue balls .Three
balls are drawn at random from the bag.What is the probability
that all of them are red ?

Sol: Let S be the sample space.
Then n(S) =number of ways drawing 3 balls out of 15.
=15 C3.
=455
Let E =event of getting all the 3 red balls.
n(E) = 5 C3 =5C2
= 10
P(E) =n(E) /n(S) =10/455 =2/91.


13)From a pack of 52 cards,one card is drawn at random.What is the
probability that the card is a 10 or a spade?
Sol: Total no of cards are 52.
These are 13 spades including tne and there are 3 more tens.
n(E) =13+3
= 16
P(E) =n(E)/n(S).
=16/52
P(E) =4/13.

14) A man and his wife appear in an interview for two vacancies in
the same post.The probability of husband's selection is 1/7 and the
probabililty of wife's selection is 1/5.What is the probabililty
that only one of them is selected?

Sol: let A =event that the husband is selected.
B = event that the wife is selected.
E = Event for only one of them is selected.
P(A) =1/7
and
p(B) =1/5.
P(A') =Probability of husband is not selected is =1-1/7=6/7
P(B') =Probaility of wife is not selected =1-1/5=4/7
P(E) =P[(A and B') or (B and A')]
= P(A and B') +P(B and A')
= P(A)P(B') + P(B)P(A')
= 1/7*4/5 + 1/5 *6/7
P(E) =4/35 +6/35=10/35 =2/7

15)one card is drawn at random from a pack of 52 cards.What is the
probability that the card drawn is a face card?

Sol: There are 52 cards,out of which there 16 face cards.
P(getting a face card) =16/52
= 4/13

16) The probability that a card drawn from a pack of 52 cards will
be a diamond or a king?

Sol: In 52 cards 13 cards are diamond including one king there are
3 more kings. E event of getting a diamond or a king.
n(E) =13 +3
= 16
P(E) =n(E) /n(S) =16/52
=4/13

17) Two cards are drawn together from apack of 52 cards.What is the
probability that one is a spade and one is a heart ?

Sol: S be the sample space the n (S) =52C2 =52*51/2
=1326
let E =event of getting 2 kings out of 4 kings
n(E) =4C2
= 6
P(E) =n(E)/n(S)
=6/1326
=1/221

18) Two cards are drawn together from a pack of 52 cards.What is the
probability that one is a spade and one is a heart?

Sol: Let S be the sample space then
n(S) =52C2
=1326
E = Event of getting 1 spade and 1 heart.
n(E) =number of ways of choosing 1 spade out of 13 and 1 heart out
of 13.
= 13C1*13C1 =169
P(E)= n(E)/n(S)
=169/1326 =13/102.

19) Two cards are drawn from a pack of 52 cards .What is the
probability that either both are Red or both are Kings?

Sol: S be the sample space.
n(S) =The number of ways for drawing 2 cards from 52 cards.
n(S) =52C2
=1326
E1 be the event of getting bothe red cards.
E2 be the event of getting both are kings.
E1nE2 =Event of getting 2 kings of red cards.
We have 26 red balls.From 26 balls we have to choose 2 balls.
n(E1) =26C2
= 26*25/2
=325
We have 4 kings .out of 4 kings,we have to choosed 2 balls.
n(E2) =4C2
=6
n(E1nE2) =2C2 =1
P(E1) = n(E1)/n(S)
=325/1326
P(E2) =n(E2)/n(S)
=6/1326
P(E1nE2) =n(E1nE2)/n(S) =1/1326
P(both red or both kings) = P(E1UE2)
= P(E1) +P(E2)-P(E1nE2)
=325/1326 +6/1326 -1/1326
=330/1326 =55/221

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2 comments:

Cindy Dy said...

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Cara
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Leslie Lim said...

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