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Showing posts with label 12th probability. Show all posts
Showing posts with label 12th probability. Show all posts

Friday, December 30, 2016

some more problems from binomial distribution for cbse ncert class xii mathematics probability

some more problems from binomial distribution for cbse ncert class xii  mathematics probability


4.Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that (i) all the five cards are spades (ii) only 3 cards are spades and (iii) none is a spade?

Let X be the number of spades among the five cards.
Assume X follows Binomial distribution with
n = 5
p = (13/52) = ( 1 / 4 ) [13 spades among the 52 cards ]
q =1 -p
q = (3 / 4)
P[X=r] = nCr prq(n-r) , r = 0,1,2,...,n
P[X=r] = 5Cr (1/4)r(3/4)(5-r) , r = 0,1,2,...,5

P[all the five cards are spades ] = P[X=5] =5C5(1/4)5(3/4)(5-5) =( 1 / 1024 )

P[only three cards are spades ] = P[X=3] =5C3(1/4)3(3/4)(5-3) =(90/1024)=(45/512)

P[none is a spade ] = P[X=0] =5C0(1/4)0(3/4)(5-0) =(243/1024)


9.On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?

Let X be the number of questions he answered correctly out of the 5 questions. just by guessing
Assume X follows Binomial distribution with
n =5
p = ( 1 / 3 ) [one out the three possible answers is correct and the candidate is guessing ]
q = 1 – p = ( 2 / 3 )
P[X=r] = nCr prq(n-r) , r = 0,1,2,...,n
P[X=r] = 5Cr ( 1 / 3 )r( 2 / 3 )(5 - r) r = 0, 1 ,..., 5
P[ candidate would get four or more correct answers just by guessing ] = P[X=4] + P[X=5]
= 5C4 ( 1 / 3 )4( 2 / 3 )(5 - 4) + 5C5 ( 1 / 3 )5( 2 / 3 )(5 - 5) = ( 11/243 )
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index of more problems on baye's theorem for ncert cbse mathematics 


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Sunday, December 25, 2016

bayes theorem problems 8 and 9 for cbse ncert

problem 8

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further,2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Let E1 be the event that the chosen item was produced by machine A .

let E2 be the event that  chosen item was produced by machine B.

let A be the event that the  item is defective .


P( E1 ) = ( 60 / 100 )

P( E2 ) = ( 40/ 100 )

P( A / E1 ) = ( 2 / 100 )

P( A / E2 ) = ( 1 / 100 )


Required probability = P [ item was produced by machine B given that the item was defective ]

Required probability = P [ E2 / A ]






P ( E2 / A ) = [( 40 / 100 )( 1 / 100 )] / { [( 60 / 100 )( 2 / 100 )] + [( 40 / 100 )( 1 / 100 )] }


P ( E2 / A ) =  [40] / {[120]+[40]} =  ( 1 / 4 )
index of more problems on baye's theorem for ncert cbse mathematics
problem 9

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Let E1 be the event that the first group wins .

let E2 be the event that the second group wins.

let A be the event that the  new product was introduced .


P( E1 ) = ( 0.6 )

P( E2 ) = ( 0.4 )

P( A / E1 ) = ( 0.7 )

P( A / E2 ) = ( 0.3 )


Required probability = P [ second group had won given that the new product was introduced ]

Required probability = P [ E2 / A ]


 
  



P ( E2 / A ) = [( 0.4 )( 0.3)] / { [( 0.6 )( 0.7 )] + [( 0.4 )( 0.3 )] }


P ( E2 / A ) =  [0.12] / {[0.42]+[0.12]} =  ( 12 / 54 ) = ( 2 / 9 )


index of more problems on baye's theorem for ncert cbse mathematics

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work 



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