Free Flip-Book Mathematics class 11th by Study Innovations. 254 Pages

Binomial Distribution.
(1) Geometrical method for probability : When the number of points in the sample space is infinite, it

becomes difficult to apply classical definition of probability. For instance, if we are interested to find the probability
that a point selected at random from the interval [1, 6] lies either in the interval [1, 2] or [5, 6], we cannot apply the
classical definition of probability. In this case we define the probability as follows:

P{x ∈ A} = Measure of region A S ,
Measure of the sample space

where measure stands for length, area or volume depending upon whether S is a one-dimensional, two-
dimensional or three-dimensional region.

(2) Probability distribution : Let S be a sample space. A random variable X is a function from the set S to
R, the set of real numbers.

{11, 12, , 16

For example, the sample space for a throw of a pair of dice is S = 21, 22, , 26
  

61, 62, , 66}

Let X be the sum of numbers on the dice. Then X(12) = 3, X(43) = 7 , etc. Also, {X = 7} is the event {61, 52,

43, 34, 25, 16}. In general, if X is a random variable defined on the sample space S and r is a real number, then
{X = r} is an event. If the random variable X takes n distinct values x1, x 2,...., xn , then {X = x1},
{X = x 2},....,{X = xn} are mutually exclusive and exhaustive events.

X = x1 X = x2 X = x3 S

X = x4 X = xn

Now, since (X = xi ) is an event, we can talk of P(X = xi ). If P(X = xi ) = Pi (1 ≤ i ≤ n) , then the system of numbers.

 x1 x2  xn 
p1 p2  pn

is said to be the probability distribution of the random variable X. The expectation (mean) of the random

n

∑variable X is defined as E(X) = pi xi
i=1

nn

∑ ∑and the variance of X is defined as var(X) = pi(xi − E(X))2 = pi xi2 − (E(X))2 .
i=1 i=1

(3) Binomial probability distribution : A random variable X which takes values 0, 1, 2, …, n is said to
follow binomial distribution if its probability distribution function is given by P(X = r) = nCr prqn−r , r = 0,1, 2,....., n

where p, q > 0 such that p + q = 1.

The notation X ~ B(n, p) is generally used to denote that the random variable X follows binomial distribution
with parameters n and p.

We have P(X = 0) + P(X = 1) + ... + P(X = n) = nC0 p0qn−0 + nC1 p1qn−1 + ... + nCn pnqn−n = (q + p)n = 1n = 1 .

Now probability of

(a) Occurrence of the event exactly r times

P(X = r) = nCr qn−r pr .
(b) Occurrence of the event at least r times

n

∑P(X ≥ r) = nCr q n−r p r + ... + p n = n C X p X q n−X .
X =r

(c) Occurrence of the event at the most r times

r

∑P(0 ≤ X ≤ r) = q n + nC1q n−1 p + ... + nCr q n−r p r = p X q n−X .
X =0

(iv) If the probability of happening of an event in one trial be p, then the probability of successive happening
of that event in r trials is pr .

Note :  If n trials constitute an experiment and the experiment is repeated N times, then the frequencies

of 0, 1, 2, …, n successes are given by N.P(X = 0), N.P(X = 1), N.P(X = 2),...., N.P(X = n).

(i) Mean and variance of the binomial distribution : The binomial probability distribution is

X0 1 2 n

P(X) nC0qnp0 nC1qn−1p nC2qn−2 p2.....nCnq0 pn

nn (npq) .

∑ ∑The mean of this distribution is Xi pi = X. nCX qn−X p X = np ,
i=1 X =1

the variance of the Binomial distribution is σ 2 = npq and the standard deviation is σ =

(ii) Use of multinomial expansion : If a die has m faces marked with the numbers 1, 2, 3, ….m and if

such n dice are thrown, then the probability that the sum of the numbers exhibited on the upper faces equal to p is

given by the coefficient of xp in the expansion of (x + x 2 + x 3 + .... + x m )n .
mn

Example: 38 A random variable X has the probability distribution :

X: 1 2 3 4 5 6 7 8

P(X) : 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the events E = {X is a prime number} and F = {X < 4}, the probability P(E ∪ F) is [AIEEE 2004]

(a) 0.50 (b) 0.77 (c) 0.35 (d) 0.87

Solution: (b) E = {X is a prime number}
Example: 39
Solution: (d) P(E) = P(2) + P(3) + P(5) + P(7) = 0.62 , F = {x < 4}

P(F) = P(1) + P(2) + P(3) = 0.50 and P(E ∩ F) = P(2) + P(3) = 0.35

∴ P(E ∪ F) = P(E) + P(F) − P(E ∩ F) = 0.62 + 0.50 − 0.35 = 0.77 .

8 coins are tossed simultaneously. The probability of getting at least 6 heads is [AISSE 1985; MNR 1985; MP PET 1994]

(a) 57 (b) 229 (c) 7 (d) 37
64 256 64 256

The required probability = 8C6  1 6. 1 2 + 8 C7  1 7 . 1  + 8C8  1 8 = 37 .
 2  2   2   2   2  256

Example: 40 An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained the probability
Solution: (a) that the minimum face value is not less than 2 and the maximum face value is not greater than 5, is

Example: 41 [IIT 1993; DCE 2000; Roorkee 2000]

Solution: (d) (a) 16 (b) 1 (c) 80 (d) 65
Example: 42 81 81 81 81
Solution: (a)
Example: 43 P(minimum face value is not less than 2 and maximum face value is not greater than 5)
Solution: (a)
= P(2 or 3 or 4 or 5) = 4 = 2 .
Example: 44 6 3
Solution: (c)
Hence required probability = 4C4  2 4  1 0 = 16 .
 3   3  81

One hundred identical coins each with probability p of showing up heads are tossed once. If 0 < p < 1 and the probability
of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is

[IIT 1988; CEE 1993; MP PET 2001]

(a) 1 (b) 49 (c) 50 (d) 51
2 101 101 101

We have 100 C50 p50(1 − p)50 = 100C51p51(1 − p)49 or 1− p = 100! 50 !. 50 ! 50 or 51 − 51p = 50p ⇒ p = 51 .
p 51!. 49 ! × 100! = 51 101

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

[AIEEE 2004]

(a) 28 (b) 219 (c) 128 (d) 37
256 256 256 256

np = 4  ⇒ q = 1 , p = 1 , n = 8
npq = 2 2 2

8  1  2  1  6 1 28
 2   2  28 256
p(X = 2) = C 2 = 28. = .

A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of
eleven steps he is one step away from the starting point is

(a) 11C6(0.24)5 (b) 11C6(0.4)6(0.6)5 (c) 11C6(0.6)6(0.4)5 (d) None of these

The man will be one step away from the starting point if (i) either he is one step ahead or (ii) one step behind the starting point.

∴ The required probability = P(i) + P(ii)

The man will be one step ahead at the end of eleven steps if he moves six step forward and five steps backward.

The probability of this event is 11C6(0.4)6(0.6)5 .
The man will be one step behind at the end of eleven steps if he moves six steps backward and five steps forward.

The probability of this event is 11C6(0.6)6(0.4)5 .

Hence the required probability = 11C6(0.4)6(0.6)5 + 11C6(0.6)6(0.4)5 = 11C6(0.4)5(0.6)5(0.4 + 0.6) = 11C6(0.24)5 .

A person can kill a bird with probability 3/4. He tries 5 times. What is the probability that he may not kill the bird

[Rajasthan PET 1997]

(a) 243/1024 (b) 781/1024 (c) 1/1024 (d) 1023/1024

Probability to kill a bird p = 3 , p+q =1
4

⇒ q = 1 − p = 1 − 3 = 1 and n=5.
4 4

Probability that he may not kill the bird,

P(X = 0) = 5C0  3 0 ⋅  1 5−0 = 1 .
 4   4  1024

Example: 45 If X follows a binomial distribution with parameters n= 8 and p = 1 , then P(| X − 4 |≤ 2) equals
Solution: (b) 2
Example: 46
Solution: (a) (a) 118 (b) 119 (c) 117 (d) None of these
128 128 128
Example: 47
Solution: (c) We have, P(| X − 4 |≤ 2) = P(−2 ≤ X − 4 ≤ 2) = P(2 ≤ X ≤ 6) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

= 8C2  1 8 + 8 C3  1 8 + 8C4  1 8 + 8C5  1 8 + 8C6  1 8 = 1 [28 + 56 + 70 + 56 + 28] = 238 = 119 .
 2   2   2   2   2  28 28 128

Three six faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is
k(3 ≤ k ≤ 8) , is

(a) (k − 1)(k − 2) (b) k(k − 1) (c) k2 (d) None of these
432 432 432

The total number of cases = 6 × 6 × 6 = 216

The number of favourable ways

= Coefficient of xk in (x + x2 + .... + x6)3

= Coefficient of xk−3 in (1 − x6)3(1 − x)−3

= Coefficient of xk−3 in (1 − x)−3 {0 ≤ k − 3 ≤ 5}

= Coefficient of xk−3 in (1 + 3C1x + 4C2x 2 + 5C3x3 + ....) = k−1C2 = (k − 1)(k − 2)
2

Thus the probability of the required event is (k − 1)(k − 2) .
432

If three dice are thrown simultaneously, then the probability of getting a score of 7 is [Kurukshetra CEE 1998]

(a) 5/216 (b) 1/6 (c) 5/72 (d) None of these

n(S) = 6 × 6 × 6

n(E) = The number of solutions of x + y + z = 7 ,

where 1 ≤ x ≤ 5,1 ≤ y ≤ 5,1 ≤ z ≤ 5

= Coefficient of x7 in (x + x 2 + .... + x 5 )3

= Coefficient of x4 in (1 + x + ..... + x 4 )3 = Coefficient of x4 in  1− x5  3
1− x

= Coefficient of x4 in (1 − 3x 5 + 3x10 − x15 )(1 − x)−3

= Coefficient of x4 in (1 − 3x 5 + 3x10 − x15 )( 2C0 + 3C1 x + 4 C2 x 2 + 5C3 x 3 + 6C4 x 4 + .....)

= 6C4 = 6! = 6×5 = 15 .
4! 2! 2

∴ p(E) = n(E) = 15 = 5 .
n(S) 6×6×6 72