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] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₅C_r (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] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₅C_r ( 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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binomial distribution problem for ncert cbse 12th mathematics probability

binomial distribution problem for ncert cbse 12th mathematics probability

1.A die is thrown 6 times. If getting an odd number is a success, what is the probability of (i) 5 successes (ii) at least 5 successes (iii) at most 5 successes?

Let X be the number of successes out of 6 throws

Assume X follows binomial distribution with

n = 6,

p = (3/6) = ( 1 / 2 )

q = 1 -p

q = ( 1 / 2 )

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₆C_r ( 1 / 2 )^r( 1 / 2 )^{(6 - r)}

P[X=r] = ₆C_r ( 1 / 2 )⁶

P[ 5 successes ] =P[X=5] = ₆C₅ ( 1 / 2 )⁶ = (3 /32)

P[ at least 5 successes ] =P[X=>5]= P[X=5] +P[X=6]= (7/64)

P[ at most 5 successes ] =P[X<=5] =1 - P[X=6]= 1 - ₆C₆ ( 1 / 2 )⁶ = (63 /64)

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2.A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of

two successes.

Let X be the number of doublets out of 4 tosses of a pair of dice.

Assume X follows binomial distribution with

n = 4,

p = (6 / 36) [6 doublets out of 36 possible outcomes in one toss of a pair of dice]

p = ( 1 / 6)

q= 1 -p

q = (5 /6)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₄C_r (1/6)^r(5/6)^{(4 - r)} , r = 0,1,2,3,4

P[two successes] = P[X = 2] =₄C₂ (1/6)²(5/6)^{(4 – 2)} = (25/216)

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This problem and the answer is from binomial distribution in the chapter on probability for class xii of cbse ncert 12^th mathematics and is useful for the students preparing for the board examination 

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finding n of binomial distribution given the probability

 finding n of binomial distribution given the probability from miscellaneous 12th cbse ncert mathematics

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Let n be the required number of times the man must toss a fair coin so that the probability of having at least one head is more than 90%

Let X be the number of heads obtained when the coin is tossed n times.

Assuming X follows binomial distribution with

n=n

p = (½)

q = 1 – p = (½)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

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hurdle problem from miscellaneous cbse ncert mathematics probability

In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is (5/6). What is the probability that he will knock down fewer than 2 hurdles?

Let X be the number of hurdles the player knocks down out of 10 hurdles

Assume X follows binomial distribution with

n=10

p = 1-(5/6) [because we defined X in terms of the hurdles knocked down ]

p=(1 /6)

q = 1 -p

q =(5/6)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

P[X=r] = ₁₀C_r (1/6)^r(5/6)^{(10 - r)} , r = 0,1,2,...,10

P[the player will knock down fewer than two hurdles] = P[ X < 2 ]

P[ X < 2 ] = P[X=0] + P[X=1]

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miscellaneous problem 4 on binomial distribution

miscellaneous problem 4 on binomial distribution for cbse xii probability

Suppose that 90% of people are right handed. What is the probability that at most 6 of a random sample of 10 people are right handed?

Let X be the number of people who are right handed out of a sample of 10 people.

Assuming X follows binomial distribution with
n = 10
p = (90/100) = (0.9) [given that 90% of people are right handed]
q = 1 - p = (0.1)

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Binomial distribution

Binomial distribution
pdf (probability distribution function ) of the binomial distribution is
Mean of the binomial distribution = np

Variance of the binomial distribution = npq

mgf (moment generating function of the binomial distribution) is [q + p (e^t) ] ^n

derivation of mean and variance of binomial distribution :derivation of mean and variance of binomial distribution

derivation of mgf of binomial distribution and deduction of mean and variance from the mgf:
mgf of the binomial distribution

derivation of mode of binomial distribution : mode of binomial distribution

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derivation of mean and variance of binomial distribution

derive the mean and variance of the binomial distribution
now to find the variance, we rewrite x^2 as x(x-1) +x before we start out



for explanation of mean of poisson distribution try the link mean of poisson distribution

for explanation of mean and variance of discrete uniform distribution try the link mean of the discrete uniform distribution



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