Binomial Distribution

As the coin is tossed 6 times the total number of outcomes will be

And we know that the favourable outcomes of getting at least 3 heads will be ⁶ + ⁶ + ⁶ + ⁶

Thus, the probability of getting at least 3 heads will be

⇒

⇒

As the coin is tossed 5 times the total number of outcomes will be .

And we know that the favourable outcomes of a head appearing even number of times will be,

That either the head appears 0, 2 or 4 times so,

The respective probabilities will be:- ⁵C₀ + ⁵C₂ + ⁵C₄ = 16

Thus, the probability

⇒

Hence, the probability is _1/2.

As 7 coins are tossed simultaneously the total number of outcomes are =128.

The favourable number of outcomes that a tail appears an odd number of times will be, ⁷C₁ + ⁷C₃ + ⁷C₅ + ⁷C₇ = 64.

Thus, the probability

Hence, the probability is _1/2 .

(i) As the coin is tossed 6 times the total number of outcomes will be = 64

And we know that the favourable outcomes of getting exactly 4 heads will be ⁶= 15

Thus, the probability of getting exactly 4 heads will be

⇒ 15/64

(ii) As the coin is tossed 6 times the total number of outcomes will be = 64

And we know that the favourable outcomes of getting at least 1 heads will be ⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ = 63

Thus, the probability of getting at least 1 head will be

⇒ 63/64

(iii) As the coin is tossed 6 times the total number of outcomes will be = 64

And we know that the favourable outcomes of getting at most 4 heads will be ⁶C₀ + ⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ = 57

Thus, the probability of getting at most 4 heads will be

⇒ 57/64

(i) As 10 coins are tossed simultaneously the total number of outcomes are =1024.

the favourable outcomes of getting exactly 3 heads will be

¹⁰C₃ = 120

Thus, the probability

Hence, the probability is .

(ii) As 10 coins are tossed simultaneously the total number of outcomes are =1024.

the favourable outcomes of getting not more than 4 heads will be

¹⁰C₀ + ¹⁰C₁ + ¹⁰C₂ + ¹⁰C₃ + ¹⁰C₄ = 386

Thus, the probability

⇒

Hence, the probability is .

(iii) As 10 coins are tossed simultaneously the total number of outcomes are =1024.

the favourable outcomes of getting at least 4 heads will be

¹⁰C₄ + ¹⁰C₅ + ¹⁰C₆ + ¹⁰C₇ + ¹⁰C₈ + ¹⁰C₉ + ¹⁰C₁₀ = 848

Thus, the probability

⇒

Hence, the probability is .

(i) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the die is thrown 6 times the total number of outcomes will be .

And we know that the favourable outcomes of getting exactly 5 successes will be, either getting 2, 4 or 6 i.e., 1/6 probability of each, total, probability, p = ,q =

The probability of success is and of failure is also .

Thus, the probability of getting exactly 5 successes will be

⇒ ⁶C₅

⇒ ⁶C_5.

⇒

(ii) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the die is thrown 6 times the total number of outcomes will be .

And we know that the favourable outcomes of getting at least 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, probability, p = , q =

The probability of success is and of failure is also .

Thus, the probability of getting at least 5 successes will be

⇒ (⁶C₅ + ⁶C₆)

⇒ (⁶C₅ + ⁶C₆)_.

⇒

(iii) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the die is thrown 6 times the total number of outcomes will be .

And we know that the favourable outcomes of getting at most 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, probability of success .

The probability of success is and of failure is also .

Thus, the probability of getting at most 5 successes will be

⇒ (⁶C₀ + ⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅).

⇒ (⁶C₀ + ⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅)_.

⇒

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

We know that the favourable outcomes of getting exactly 3 successes will be, either getting 1 or a 6 i.e, total, probability

The probability of success is and of failure is .

Thus, the probability of getting exactly 3 successes will be

⇒ (⁴C₃)

⇒ (⁴C₃)_.

⇒

(ii) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

We know that the favourable outcomes of getting at least 2 successes will be, either getting 1 or a 6 i.e, total, probability

The probability of success is and of failure is .

Thus, the probability of getting at least 2 successes will be

⇒ (⁴C₂) ) + (⁴C₃) ) + (⁴C₄)

⇒

⇒

(iii) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

We know that the favourable outcomes of getting at most 2 successes will be, either getting 1 or a 6 i.e, total, probability

The probability of success is and of failure is .

Thus, the probability of getting at most 2 successes will be

⇒ (⁴C₀) + (⁴C₁) + (⁴C₂)

⇒

⇒

The total outcomes = 36,

The favourable outcomes are (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3) ,(4,5), (4,6)

Thus, the probability = favourable outcomes/total outcomes

⇒

As the pair of die is thrown 4 times,

The total number of outcomes = 36

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

The probability of success = p =

q =

probability of 2 successes = ⁴C₂.()²()²

⇒

(i) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =7

the favourable outcomes ,

(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)

The probability of success = p =

q =

probability of no success = ⁷C₀.()⁰()⁷

⇒ ()⁷

(ii) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =7

the favourable outcomes ,

(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)

The probability of success = p =

q =

probability of exactly 6 successes = ⁷C₆.()⁶()¹

⇒ 35.()⁷

(iii) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =7

the favourable outcomes ,

(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)

The probability of success = p =

q =

probability of at least 6 successes =

⁷C₆.()⁶()¹ + ⁷C₇.()⁷()⁰

⇒ 36.()⁷

⇒ ()⁵

(iv) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =7

the favourable outcomes ,

(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)

The probability of success = p =

q =

probability of at least 6 successes =

⁷C₀.()⁰()⁷ + ⁷C₁.()¹()⁶ + ⁷C₂.()²()⁵ + ⁷C₃.()³()⁴ + ⁷C₄.()⁴()³ + ⁷C₅.()⁵()² + ⁷C₆.()⁶()¹

⇒ (1- ⁷)

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =8

The probability of success, i.e. the bulb is defective = p =

q =

probability of that there is not more than one defective piece=

P(0 defective items) + P(1 defective item) =

⁸C₀.()⁰()⁸ + ⁸C₁.()¹()⁷

⇒ (⁷)

(i) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =5

The probability of success, i.e. the bulb is defective = p =

q =

probability of that no bulb is defective piece=

P(0 defective items) =

⁵C₀.()⁰()⁵

⇒ (⁵

(ii) Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =5

The probability of success, i.e. the bulb is defective = p =

q =

probability of that there are exactly 2 defective pieces=

P(2 defective items) =

⁵C₂.()²()³

⇒ (

(i) The probability that the bulb will fuse = 0.05 = p

The probability that the bulb will not fuse = 1-0.05 = 0.95 = q

Using Bernoulli’s we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =5

Probability that none will fuse =

⁵C₀.(0.05)⁰(0.95)⁵

⇒ (0.95)⁵

(ii) The probability that the bulb will fuse = 0.05 = p

The probability that the bulb will not fuse = 1-0.05 = 0.95 = q

Using Bernoulli’s we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =5

Probability that at least one will fuse = P(1) + P(2) + P(3) + P(4) + P(5)

⁵C₁.(0.05)¹(0.95)⁴ + ⁵C₂.(0.05)²(0.95)³ + ⁵C₃.(0.05)³(0.95)² + ⁵C₄.(0.05)⁴(0.95)¹ + ⁵C₅.(0.05)⁵(0.95)⁰

⇒ (1-(0.95)⁵)

(iii) The probability that the bulb will fuse = 0.05 = p

The probability that the bulb will not fuse = 1-0.05 = 0.95 = q

Using Bernoulli’s we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =5

Probability that not more than one will fuse = P(0) + P(1)

⁵C₀.(0.05)⁰(0.95)⁵ + ⁵C₁.(0.05)¹(0.95)⁴

⇒ (1.20).(0.95)⁵

(i) The probability that the item is defective = = p

The probability that the bulb will not fuse = 1= = q

Using Bernoulli’s we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

The probability that exactly 2 defective items are,

⇒ ⁶C₂.()²()⁴

⇒

(ii) The probability that the item is defective = = p

The probability that the bulb will not fuse = 1= = q

Using Bernoulli’s we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

The probability that not more than 2 defective items are,

⇒ ⁶C₀.()⁰()⁶ + ⁶C₁.()¹()⁵ + ⁶C₂.()²()⁴

⇒

(iii) The probability that the item is defective = = p

The probability that the bulb will not fuse = 1= = q

Using Bernoulli’s we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

The probability of at least 3 defective items are,

P(3) + P(4) + P(5) + P(6)

⇒ ⁶C₃.()³()³ + ⁶C₄.()⁴()² + ⁶C₅.()⁵()¹ + ⁶C₆.()⁶()⁰

⇒

The probability that the called number is busy is

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

The probability that at least three of them will be busy is:-

P(0) + P(1) + P(2) + P(3)

⇒ ⁶C₀()⁰()⁶ + ⁶C₁()¹()⁵ + ⁶C₂()²()⁴ + ⁶C₃()³()³

⇒ 1 – ()⁴.()

The probability that any one of them has an accident is 0.1.

The probability any car reaches safely is 0.9.

The probability that all the cars reach the finishing line without any accident is = (0.9)(0.9)(0.9) = 0.729

The probability that the operations performed are successful is = 0.8

The probability that at least three operations are successful is = P(3) + P(4)

⇒ ⁴C₃(0.8)³(0.2)¹ + ⁴C₄(0.8)⁴(0.2)⁰

⇒

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =7

p = � q = �

The probability of hitting the target at least twice is = P(2) + P(3) + P(4) + P(5) + P(6) + P(7)

⇒ 1-(P(0) + P(1))

⇒ 1 – (⁷C₀()⁰()⁷ + ⁷C₁()¹()⁶)

⇒ 1 – ()( ()⁶)

⇒

The probability that the hurdle will be cleared is 5/6

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =10

p = 5/6 q = 1/6

Probability that he will knock down fewer than 2 hurdles is =

P(0) + P(1)

⇒ ¹⁰C₀()⁰()¹⁰ + ¹⁰C₁()¹()⁹

⇒

The probability that the bird will be shot, is 1/3

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =3

p = 1/3 q =2/3

Probability that he will hit at least one bird is =

P(1) + P(2) + P(3)

⇒ ³C₁()¹()² + ³C₂()²()¹ + ³C₃()³()⁰

⇒

The probability that a man aged 60 will live to be 70 is 0.65

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =8

p = 0.65 q =0.35

Probability that out of 10 men, now 60, at least 8 will live to be 70 is: P(8) + P(9) + P(10)

¹⁰C₈(0.65)⁸(0.35)² + ¹⁰C₉(0.65)⁹(0.35)¹ + ¹⁰C₁₀(0.65)¹⁰(0.35)⁰

⇒ 0.2615

(i) Balls are drawn at random,

So, the probability that none is white is,

In a trial the probability of selecting a non-white ball is

So, in 4 trials it will be,

⇒ ()()()()=

(ii) Balls are drawn at random,

So, the probability that all are white is,

In a trial the probability of selecting a white ball is

So, in 4 trials it will be,

⇒ ()()()()=

(iii) Balls are drawn at random,

So, the probability that at least one is white is,

In a trial the probability of selecting a white ball is

So, in 4 trials the probability that at least one is white is,

Selecting a white and then choosing from the rest,

⇒ 1- that no ball is white

is .

The probability that the burglar will be hit by a bullet is 0.6.

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =6

p = 0.6 q =0.4

The probability that the burglar is unhurt is,

⁶C₀(0.6)⁰(0.4)⁶

⇒ 0.004096

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =3

p = 2/6 = 1/3, q = 4/6 = 2/3

P(x = 0) = P (no success) = P (all failures) = ()()()=

P(x = 1) = P (1 success and 2 failures) = ³C₁()¹()² =

P(x = 2) = P (2 success and 1 failure) =³C₂()²()¹ =

P(x = 3) = P (all 3 success) = ³C₃()³()⁰ =

∴ The probability distribution of the random variable x is -

x : 0 1 2 3

P(x) :

x₁ p₁ p₁x₁ p₁x₁²

0 0 0

1

2

3

1

Mean μ = Σp₁x₁ = 1

Variance = ² = Σp₁x₁² - μ

⇒ 5/3 - 1/1

⇒ 2/3

Probability of getting an even number is = 3/6 = 1/2

Probability of getting an odd number is = 3/6 = 1/2

Variance = npq

⇒ 100

⇒ 25

Mean = np = 9

Variance = npq = 6

⇒ q = =

⇒ p = 1 - =

⇒ n = 27

Binomial distribution

where r = 0, 1, 2, 3, ………, 27

Mean = np = 5

Variance = npq = 2.5

⇒ q = =

⇒ p = 1 - =

⇒ n = 10

Probability distribution is:-

Mean = np = 4

Variance = npq = 4/3

⇒ q =

⇒ p = 1 - =

⇒ n = 6

The probability (X) is

⁶C₁()¹()⁵ + ⁶C₁()¹()⁵ + ⁶C₁()¹()⁵ + ⁶C₁()¹()⁵ + ⁶C₁()¹()⁵ + ⁶C₁()¹()⁵

=

Mean = np = 6

Variance = npq = 2

⇒ q =

⇒ p = 1 - =

⇒ n = 9

The probability of getting 5 successes,

⁹C₅()⁵()⁴

Mean + Variance = np + npq = np(1 + q) = 25/3

Variance = n²p²q = n2 = 50/3 …(i)

n²p²(1 + q)² = 625/9 …(ii)

Dividing (i) by (ii), we get,

⇒ 6q2-13q + 6 = 0

⇒ q = 2/3 or 3/2

⇒ But as q can not be greater than 1 thus, q = 2/3.

⇒ p = 1/3

⇒ n = 15

Binomial distribution,

Mean is 10,

Standard deviation is

So, variance is ² i.e. 8

Thus,

Mean = np = 10

Variance = npq = 8

⇒ q =

⇒ p = 1 - =

⇒ n = 50

Thus, the binomial distribution is

.

Variance can not be greater than mean as then, q wll be greater than 1, which is not possible.

As, np = 6 and npq = 9

q = 3/2 …(not possible)

If A and B are mutually exclusive events then,

P(A) = 0.4, P(B) = X

And P(A B) = P(A)+P(B) = 0.5 = 0.4 + P(B)

⇒ P(B) = 0.1

As A and B are independent events such that P(A) = 0.4, P(B) = x

So, P(AB) = P(A)P(B)

And P(A B) = P(A)+P(B)+P(AB)

P(A B) = 0.4 + X - 0.4X = 0.5

⇒ 0.4 + 0.6X = 0.5

⇒ X = 1/6

P(B/A) =

⇒ And P(A) = 0.8,

⇒ P() = 0.32

So, P(A/B) =

⇒ P(A/B) = = 0.64

⇒ Hence, the answer is b.

P(A) = , P(B) = and P(A ∪ B) =

P(A ∪ B) = P(A) + P(B) - P()

⇒ P()

⇒ P(AB) =

P(A/B) = P(A B)/P(B)

_⇒ P(A/B) = =

We are having two events A and B such that

P(A) = , P(B) = and P(A’ ∪ B’) = ,

P(A’∪ B’) = P’(A B) = 1 – P(A B) =

⇒ P(A B) =

⇒ As P(A B) P(A).P(B) … thus, they are not independent,

⇒ And as P(A ∪ B) P(A) + P(B) … thus, they are not mutually exclusive.

Hence, the answer is option d.

_{P(A) = probability that A can solve the problem}

_{= 3/5}

_{And P(B) = probability that B can solve the problem = 2/3}

_{P(A U B) = P(A) + P(B), As the events are independent}

_{⇒P(A∩B) = P(A).P(B)}

_Thus,

⇒ _{P(A) + P(B) = 3/5 +2/3 – 2/5 = 13/15}

The probability that the problem is solved = P(AB) = P(A)+P(B)+P(C)-P(AB)-P(BC)-P(CA) + 3P(ABC)

Considering independent events, P(AB) =P(A).P(B),

P(BC) =P(B).P(C), P(CA) =P(C).P(A),

P(ABC) =P(A).P(B).P(C),

Thus, P(AB) is,

⇒

P(A) = P(B) = P(C) =

P(B CA’) = P(B C) – P(B CA)

As the events are independent, So, P(B C) = P(B).P(C) =

And P(B CA) = P(B).P(C).P(A) =

P(B CA’) =

The probability of failure of the first component = 0.2 =P(A)

The probability of failure of second component = 0.3 = P(B)

The probability of failure of third component = 0.5 = P(C)

As the events are independent,

The machine will operate only when all the components work, i.e.,

(1-0.2)(1-0.3)(1-0.5) = P(A’)P(B’)P(C’)

In rest of the cases, it won’t work,

So P(AUBUC) = 1 – P(A’B’C’) = 1 – (0.8).(0.7).(0.5)

⇒ 1 – 0.28 = 0.72

The probability that the outcome which is either, 1, 3 or 5 is prime is

=

Favourable outcomes = 3 or 5

Total outcomes = 1, 3, and 5

Thus, probability=

⇒

P(A) = 0.3, P(B) = 0.2 and P(A ∩ B) = 0.1

P(∩ B) = P(B) – P(A ∩ B) = 0.2 – 0.1 = 0.1

P(A) = , P(B) = and P(A ∩ B) = ,

P(=

⇒ P( =

P(A) = 0.4, P(B) = 0.8 and

P(B/A) = 0.6,

P(B/A) =

P(A B) = 0.24

⇒ P(A/B) =

P =

Given,

P(A ∪ B) = , P(A ∩ B) = and

P() = , P(B) = 1- P() = 1- =

⇒ P(B) =

⇒ P(A) =

⇒ Hence, these are independent.

The die is thrown twice,

So the favourable outcomes that the sum appears to be 7 are

(1,6), (2,5), (3,4), (4,3), (5,2) and (6,1)

Out of these 2 appears twice,

So the probability that 2 appears at least once is:

=

⇒

_{The sum will be even when; both numbers are either even or odd,}

_{i.e. for both numbers to be even, the total cases}⁵C₁X⁴C₁ (Both the numbers are odd)+⁴C₁X³C₁ (Both the numbers are even)= 32

_{The favourable number of cases will be,}

_{Both odd, i.e. selecting numbers from 1, 3, 5, 7, or 9, i.e.}

_⁵C1X⁴C₁ = 20

_{Thus, the probability that both numbers are odd will be =}

=

⇒

_Given:

60% of the students read mathematics, 25% biology and 15% both mathematics and biology

That means,

Let the event A implies students reading mathematics,

Let the event B implies students reading biology,

Then, P(A) = 0.6

P(B) = 0.25

P(AB) = 0.15

We, need to find P(A/B) = P(AB)/ P(B)

⇒

_{The couple has two children and one is known to be boy,}

_{The probability that the other is boy will be =}

_{Total outcomes are 3,}

_{The first child is a boy, the second girl}

_{The first child is a girl, the second boy}

_{The first child is a boy, second boy}

_{The favourable outcome is one,}

_{Thus, the probability that the other is boy will be}

_⇒1/3

_{A die is tossed twice,}

_{The probability of getting a 4, 5 or 6 in the first trial is 3/6 = P(A)}

_{The probability of getting a 1, 2, 3 or 4 in the second trial is 4/6 =P(B)}

_{As the events are independent, the probability of these two events together will be, P(A).P(B) = 1/3.}

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the coin is thrown 6 times the total number of outcomes will be .

And we know that the favourable outcomes of getting at least 3 successes will be, getting a head

The probability of success is and of failure is also

⁶C₃()³()³ +⁶C₄()⁴()² +⁶C₅ ()⁵()¹ + ⁶C₆()⁶()⁰

⇒

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the coin is tossed 5 times the total number of outcomes will be .

And we know that the favourable outcomes of getting the odd tail number of times ,successes will be, getting a tail

The probability of success is and of failure is also

⁵C₁()¹()⁴ +⁵C₃()³()² +⁵C₅ ()⁵()⁰

⇒

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the coin is tossed 5 times the total number of outcomes will be .

And we know that the favourable outcomes of getting the head even number of times ,successes will be, getting a head,

The probability of success is and of failure is also

the probability that head appears an even number of times =

P(0)+P(2)+P(4)

= ⁵C₂()²()³ +⁵C₃()³()² +⁵C₅ ()⁵()⁰

⇒

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the coin is tossed 8 times the total number of outcomes will be .

And we know that the favourable outcomes of getting at least 6 heads are,successes will be, getting a head,

The probability of success is and of failure is also

the probability of getting at least 6 heads is =

P(6) +P(7) +P(8)

= ⁸C₆()⁶()² +⁸C₇()⁷()¹ +⁸C₈()⁸()⁰

⇒

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As the die is thrown 5 times the total number of outcomes will be .

And we know that the favourable outcomes of getting at least 4 successes will be, either getting 1, 3 or 5 i.e., 1/6 probability of each, total, probability, p = ,q =

The probability of success is and of failure is also

the probability of getting at least 4 successes =

P(4)+P(5)

⇒ ⁵C₄()⁴()¹ +⁵C₅()⁵()⁰

⇒

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p)

As we know that the favourable outcomes of getting at least doublets twice are, successes will be, getting a doublet, i.e.,

, p = ,q =

The probability of success is and of failure is also

the probability of getting at least 2 successes =

P(2)+P(3)+P(4)

⇒ ⁴C₂()²()² +⁴C₃()³()¹+⁴C₄()⁴()⁰

⇒

Using Bernoulli’s Trial P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), here n = 7

As we know that the favourable outcomes of getting at most 6 success are, successes will be, getting a total of 7 is success, i.e.,

We can get 7 by, (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

, p = ,q =

The probability of success is and of failure is also

the probability of getting at most 6 successes =

P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1-P(7)

⇒ 1 – ⁷C₇()⁷()⁰

⇒ 1 - ()⁷

The probability that the man hits the target is 3/4

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =5

p = �, q = �

Probability that he will hit at least 3 times is =

P(3)+P(4)+P(5)

⇒ ⁵C₃()³()²+⁵C₄()⁴()¹+⁵C₅()⁵()⁰

⇒

The probability of safe arrival of the ship is 1/5

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =5

p = 1/5 , q = 4/5

Probability of safe arrival of at least 3 ships is =

P(3)+P(4)+P(5)

⇒ ⁵C₃()³()²+⁵C₄()⁴()¹+⁵C₅()⁵()⁰

⇒

The probability of occurrence of an event E in one trial is 0.4

Using Bernoulli’s Trial we have,

P(Success=x) = ⁿC_x.p^x.q^(n-x)

x=0, 1, 2, ………n and q = (1-p), n =3

p = 0.4 , q = 0.6

The probability that E occurs at least once is,

P(1)+P(2)+P(3)

⇒ ³C₁()¹()²+³C₂()²()¹+³C₃()³()⁰

⇒