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**LONG ANSWER TYPE QUESTIONS**

**Q.41. Three bags contain a number of red and white balls as follows:Bag 1 : 3 red balls,Bag 2 : 2 red balls and 1 white ballBag 3 : 3 white balls.**

(1) a red ball will be selected?

(2) a white ball is selected?

Bag I: 3 red balls and no white ball

Bag II: 2 red balls and 1 white ball

Bag III: no red ball and 3 white balls Let E

(i) Let E be the event that red ball is selected

âˆ´ P(E) = P(E

(ii) Let F be the event that a white ball is selected

âˆ´ P(F) = 1 â€“ P(E)

[P(E) + P(F) = 1]

Hence, the required probabilities are 7/18 and 11/8.

Hence, the required probabilities are 2/11 and 9/11.

(3) That it is of the type A

where A

Let E be the event that a seed germinates andbe the event that a seed does not germinate

(iii) Using Bayesâ€™ Theorem, we get

Hence, the required probability is 16/51 or 0.314.

Let E

and E

Also E

[âˆµ For TATA NAGAR, the two consecutive letters visible are TA, AT, TA, AN, NA, AG, GA, AR]

âˆ´ P (E

Now using Bayesâ€™ Theorem, we have

Hence, the required probability is 7/11.

and E

Let E

Hence, the required probability is 11/21.

âˆ´ Probabilities of choosing either of the urns are

P(U

Let H be the event of drawing white ball from the chosen urn.

Hence, the required probability is 1/3.

Let E

E

and H : Event that the person is diagnosed to have TB.

Hence, the required probability is 110/221.

Let E

E

E

Let H be the event that the selected item is defective.

âˆ´ Using Bayesâ€™ Theorem,

Hence, the required probability is 5/11.

where k is a constant. Calculate

(i) The value of k

(ii) E (X)

(iii) Standard deviation of X.

(i) Here, P(X = x) = k(x + 1) for x = 1, 2, 3, 4

So, P(X = 1) = k(1 + 1) = 2k; P(X = 2) = k(2 + 1) = 3k

P(X = 3) = k(3 + 1) = 4k; P(X = 4) = k(4 + 1) = 5k

Also, P(X = x) = 2kx for x = 5, 6, 7

P(X = 5) = 2(5)k = 10k; P(X = 6) = 2(6)k = 12k

P(X = 7) = 2(7)k = 14k

and for otherwise it is 0.

âˆ´ The probability distribution is given by

We know that

So, 2 k + 3k + 4k + 5k + 10k + 12k + 14k = 1

â‡’ 50k = 1

â‡’ k = 1/50

Hence, the value of k is 1/50

(ii) Now the probability distribution is

(ii) We know that Standard deviation (SD) =

Variance = E(X

âˆ´ Variance (X)

âˆ´ S.D = âˆš2.92 = 1.7 (approx.)

Calculate :

(i) The value of A if E(X) = 2.94

(ii) Variance of X.

âˆ´

= 0.5 + 0.8 + 1.92 + 3.6 + 3.24 + 9.00 = 19.06

Variance (X) = E(X

= 19.06 â€“ (2.94)

where k is a constant. Calculate

(1) E(X)

(2) E (3X

(3) P(X â‰¥ 4)

Given that:

P(X = x) =

âˆ´ Probability distribution of random variable X is

We know that

âˆ´ k + 4k + 9k + 8k + 10k + 12k = 1

â‡’ 44k = 1

â‡’ k = 1/44

Ans.

Given that n coins are two headed coins and the remaining (n + 1) coins are fair.

Let E

E

E : the event that the toss results in a head

âˆ´

P(E/E1) = 1 (sure event) and P (E/E

âˆ´

Hence, the required value of n is 10.

Ans.

and E = the event of drawing an ace

and F = the event of drawing non-ace.

We have Distribution Table:

Now, Mean E(X) =

âˆ´ Variance = E(X

Standard deviation =

= 0.377(approx.)**Q.54. A die is tossed twice. A â€˜successâ€™ is getting an even number on a toss. Find the variance of the number of successes.Ans. **

Let E be the event of getting even number on tossing a die.

Here X = 0, 1, 2

âˆ´ Probability distribution table is

âˆ´ Variance (X) = E(X

=

Ans.

âˆ´ n(S) = 20

Let X be the random variable denoting the sum of the numbers on two cards drawn.

âˆ´ X = 3, 4, 5, 6, 7, 8, 9

So,

âˆ´Variance (X) = E(X

**OBJECTIVE TYPE QUESTIONS**

**Choose the correct answer from the given four options in each of the exercises from 56 to 82.Q.56. If P(A) = 4/5 and P(A âˆ© B) = 7/10 then P (B/A) is equal to(1) 1/10(2) 1/8(3) 7/8(4) 17/20Ans. **

Given that: P(A) = 4/5 and P(A âˆ© B) = 7/10

âˆ´

Hence, the correct option is (3).

(1) 14/17

(2) 17/20

(3) 7/8

(4) 1/8

Ans.

Given that: P(A âˆ© B) = 7/10 and P(B) = 17/20

âˆ´

Hence, the correct option is (1).

(1) 1/4

(2) 1/3

(3) 5/12

(4) 7/2

Ans.

Here, P(A) = 3/10, P(B) = 2/5 and P(A âˆª B) = 3/5

P(A âˆª B) = P(A) + P(B) - P(A âˆ© B)

3/5 = 3/10 + 2/5 - P(A âˆ© B)

Now P (A/B) +P (B/A) =

Hence, the correct option is (4).

(1) 5/6

(2) 5/7

(3) 25/42

(4) 1

Ans.

Given that: P(A) = 2/5, P(B) = 3/10 and P(A âˆ© B) = 1/5

and P(A' âˆ© B') = 1 - P(A âˆª B) = 1 - [P(A) + P(B) - P(A âˆ© B)]

âˆ´

and

Hence, the correct option is (3).

(1) 1/12

(2) 3/4

(3) 1/4

(4) 3/16

Ans.

Given that: P(A) = 1/2 ,P(B) = 1/3 and P (A/B) = 1/4

Now P(A' âˆ© B') = 1 - P(A âˆª B)

= 1 - [P(A) + P(B) - P(A âˆ© B)]

Hence, the correct option is (3).

Q.61. If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6, then P(A âˆª B) is equal to

(1) 0.24

(2) 0.3

(3) 0.48

(4) 0.96

Given that: P(A) = 0.4, P(B) = 0.8 and P (B/A) = 0.6

âˆ´ P(A âˆ© B) = 0.6 Ã— 0.4 = 0.24

P(A âˆª B) = P(A) + P(B) - P(A âˆ© B)

= 0.4 + 0.8 â€“ 0.24 = 1.20 â€“ 0.24 = 0.96

Hence, the correct option is (4).

(1) P(A|B) = P(A).P(B)

(2) P(A|B) = P(A âˆ© B)/P(B)

(3) P(A|B).P(B|A)=1

(4) P(A|B) = P(A)|P(B)

Ans.

Given that: A = Ï† and B â‰ Ï†,

then

Hence, the correct option is (2).

(1) 2/3

(2) 1/2

(3) 3/10

(4) 1/5

Ans.

Given that: P(A) = 0.4, P(B) = 0.3 and P(A âˆª B) = 0.5

P(A âˆª B) = P(A) + P(B) - P(A âˆ© B)

0.5 = 0.4 + 0.3 â€“ P(A âˆ© B)

P(A âˆ© B) = 0.4 + 0.3 - 0.5 = 0.2

âˆ´ P(B' âˆ© A) = P(A) - P(A âˆ© B)

= 0.4 â€“ 0.2 = 0.2 = 1/5

Hence, the correct option is (4).

(1) 3/10

(2) 1/5

(3) 1/2

(4) 3/5

Ans.

Given that: P(B) = 3/5, P (A/B) = 1/2 and P(A âˆª B) = 4/5

We know that P (A/B) = P(A âˆ© B)/P(B)

1/2 =

âˆ´ P(A âˆ© B) = 3/10

Now P(A âˆª B) = P(A) + P(B) - P(A âˆ© B)

Hence, the correct option is (3).

(1) 1/5

(2) 3/10

(3) 1/2

(4) 3/5

Ans.

Hence, the correct option is (4).

(1) 1/5

(2) 4/5

(3) 1/2

(4) 1

Ans.

âˆ´

Now P(A âˆª B)' + P(A' âˆªB)

Hence, the correct option is (4).

(1) 6/13

(2) 4/13

(3) 4/9

(4) 5/9

Ans.

Given that: P(A) = 7/13, P(B) = 9/13 and P(A âˆ© B) = 4/13

(1) 1 â€“ P(A | B)

(2) 1â€“ P( A â€² | B)

(3)

(4) P(Aâ€²) | P( Bâ€²)

Ans.

Given that: P(A) > 0 and P(B) â‰ 1

âˆ´

Hence, the correct option is (3).

(1) 4/15

(2) 8/45

(3) 1/3

(4) 2/9

Ans.

such that

âˆ´

Hence, the correct option is (d).

(1) They must be mutually exclusive

(2) The sum of their probabilities must be equal to 1

(3) (1) and (2) both are correct

(4) None of the above is correct

Ans.

So, they will not be mutually exclusive.

If P(A) + P(B) = 1, they are exhaustive events and for independent events A and

P(A âˆ© B) â‰ 0.

Hence, the correct option is (4).

(1) 2/5

(2) 3/8

(3) 3/20

(4) 6/25

Ans.

Now P (A/B) .P (A'/B)

Hence, the correct option is (4).

(2) P(A) â€“ P(B)

(3) P (A) . P(B)

(4) P(A) | P(B)

Since A and B are two independent events

âˆ´ P(A âˆ© B) = P(A).P(B)

Hence, the correct option is (3).

(1) 2/7

(2) 3/35

(3) 1/70

(4) 1/7

Ans.

Given that: E and F are independent events such that

Hence, the correct option is (3).

(1) 45/196

(2) 135/392

(3) 15/56

(4) 15/29

Ans.

Given that: Bag contains 5 red and 3 blue balls.

Probability of getting exactly one red ball if 3 balls are randomly drawn without replacement

= P(R) . P(B) . P(B) + P(B) . P(R) . P(B) + P(B) . P(B) . P(R)

Hence, the correct option is (3).

(1) 1/3

(2) 4/7

(3) 15/28

(4) 5/28

Ans.

According to Question 74,

Let E

E

âˆ´ P(E

P(E

Hence, the correct option is (2).

(1) 0.024

(2) 0.188

(3) 0.336

(4) 0.452

Ans.

Given that: P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2

Also = 1 â€“ 0.4 = 0.6,= 1 â€“ 0.3 = 0.7

and = 1 â€“ 0.2 = 0.8

âˆ´ Probabilities of two hits

Hence, the correct option is (2).**Q.77. Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is(1) 1/2(2) 1/3(3) 2/3(4) 4/7Ans.**

Let G denotes the girl and B denotes the boy of the given family.

So, n(S) = {(BGG), (GBG), (GGB), (GBB), (BGB), (BBG), ( BBB), ( GGG)}

Let E

âˆ´ E

Let E

âˆ´E

(E

Hence, the correct option is (4).

(1) 1/2

(2) 1/4

(3) 1/8

(4) 3/4

Ans.

Let E

So,

Hence, the correction option is (3).

(1) 3/28

(2) 2/21

(3) 1/28

(4) 167/168

Ans.

= P(G).P(G).P(B) + P(G) . P(B).P(G) + P(B).P(G).P(G)

Hence, the correct option is (1).

(1) 33/56

(2) 9/64

(3) 1/14

(4) 3/28

Ans.

Required probability = P(dead).P(dead)

Hence, the correct option is (4)

(1) 1/256

(2) 7/32

(3) 5/32

(4) 3/32

Ans.

Here, n = 8, p = 1/2, q = 1-(1/2) = 1/2 and r = 3

we know that

Hence, the correct option is (2).

(1) 1/18

(2) 5/18

(3) 1/5

(4) 2/5

Ans.

Let E

âˆ´ E

n(E

and E

â‡’ n(E

âˆ´ Required probability

Hence, the correct option is (3).

We know that for a Binomial distribution, the outcomes must not be dependent on each other.

Hence, the correct option is (3).

**(1)(2)(3)**

Ans.

Probability of getting Queen = 4/52

So, the required probability = P(Queen).P(Queen)

(with replacement)

Hence, the correct option is (1).

(1) 7/64

(2) 7/128

(3) 45/1024

(4) 7/41

Ans.

Here, n = 10, p = 1/2 and q = 1/2 (for true/false questions)

and r â‰¥ 8 i.e. 8, 9, 10

âˆ´ P(X â‰¥ 8) = P(x = 8) + P(x = 9) + P(x = 10)

Hence, the correct option is (2).

(1)

(2)

(3)

(4) (0.7)

Ans.

Given that:

âˆ´ p = 0.7 and q = 1 â€“ 0.7 = 0.3

n = 5 and r = 4

We know that

Hence, the correct option is (1).

X | 2 | 3 | 4 | 5 |

P(X) | 5/k | 7/k | 9/k | 11/k |

**The value of k is(1) 8(2) 16(3) 32(4) 48Ans.** We know that

Hence, the correct option is (3).

X | -4 | -3 | -2 | -1 | 0 |

P(X) | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |

**E(X) is equal to :(1) 0(2) â€“1(3) â€“2(4) â€“1.8Ans. **We know that:

= (â€“ 4)(0.1) + (â€“ 3)(0.2) + (â€“ 2)(0.3) + (â€“ 1)(0.2) + 0(0.2)

= â€“ 0.4 â€“ 0.6 â€“ 0.6 â€“ 0.2 = â€“ 1.8

Hence, the correct option is (d).

X | 1 | 2 | 3 | 4 |

P(X) | 1/10 | 1/5 | 3/10 | 2/5 |

**E(X ^{2}) is equal to(1) 3(2) 5(3) 7(4) 10Ans.**We know that

Hence, the correct option is (4).

(1) 1/2

(2) 1/3

(3) 1/5

(4) 1/7

Ans.

Now

The above expression will be independent of n and r if

Hence, the correct option is (1).

(1) 1/10

(2) 2/5

(3) 9/20

(4) 1/3

Ans.

Let E

Hence, the correct option is (2).

(1) 1/12

(2) 1/40

(3) 13/120

(4) 10/13

Ans.

Let E

and E

Let H be the event that both of them get the same answer.

Here, P (H/E

Hence, the correct option is (4).

(A)

(B)

(C)

(D)

Ans.

Here, n = 5, p = 10/100 = 1/10 and q = 1-(1/10) = 9/10 and r â‰¤ 1

we know that

Hence, the correct option is (4).

Q.94. Let P(A) > 0 and P(B) > 0. Then A and B can be both mutually exclusive and independent.

Ans.

Q.95. If A and B are independent events, then A â€² and Bâ€² are also independent.

Ans.

Q.96. If A and B are mutually exclusive events, then they will be independent also.

Ans.

Q.97. Two independent events are always mutually exclusive.

Ans.

Q.98. If A and B are two independent events then P(A and B) = P(A).P(B).

Ans.

Q.99. Another name for the mean of a probability distribution is expected value.

Ans.

Ans.

P (exactly one of A, B occurs) = P (A) P (Bâ€²) + P ( B) P ( Aâ€² )

Ans.

Ans.

Ans.

and

Ans.

Given that: P(A) = p, P(B) = 1/3 and P(A âˆª B) = 5/9

Hence, p is equal to 1/3 .

then P(A') + P(B') =

Ans.

Here,

âˆ´

Now P(A') + P(B') = 1 - P(A) + 1 - P(B) = 2 - [P(A) + P(B)]

= 2 - [P(A âˆª B) + P(A âˆ© B)]

Hence, the value of the filler is 10/9.

Ans.

Given that: P(X = 2) = 9P(X = 3)

[âˆµ

â‡’ 9p = q

â‡’ 9p = 1 â€“ p

â‡’ 9p + p = 1

â‡’ 10p = 1

âˆ´ p = 1/10

Hence, the value of the filler is 1/10 .

Ans.

Hence, Var (X) is equal to

Ans.

So, A is independent of B.

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