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**SHORT ANSWER TYPE QUESTIONS**

**Q.1. For a loaded die, the probabilities of outcomes are given as under:P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3.The die is thrown two times. Let A and B be the events, â€˜same number each timeâ€™, and â€˜a total score is 10 or moreâ€™, respectively. Determine whether or not A and B are independent.Ans. **A loaded die is thrown such that

P(1) = P(2) = 0.2,

P(3) = P(5) = P(6) = 0.1

and P(4) = 0.3

die is thrown two times.

Also given that:

A = Same number each time and

B = Total score is 10 or more.

So, P(A) - [P(1,1) + P(2, 2) + P(3, 3) + P(4, 4) + P(5, 5) + P(6, 6)]

= P(1).P(1) + P(2).P(2) + P(3).P(3) + P(4).P(4) + P(5).P(5) + P(6).P(6)

= 0.2 x 0.2 + 0.2 x 0.2 + 0.1 x 0.1 + 0.3 x 0.3 + 0.1 x 0.1 + 0.1 x 0.1

= 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20

Now B = ((4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)]

P(B) = |P(4).P(6) + P(6).P(4) > P(5).P(5) + P(5).P(6) + P(6).P(5) + P(6).P(6)

= 0.3 x 0.1 +0.1 x 0.3+ 0.1 x 0.1 +0.1 x 0.1 +0.1 x 0.1 + 0.1 x 0.1

= 0.03 + 0.03 + 0.01 + 0.01 + 0.01 + 0.01 = 0.10

A and B both events will be independent if

P(A âˆ© B) = P(A).P(B) ...(i)

Here, (A âˆ© B) = {(5, 5), (6,6)}

âˆ´ P(A âˆ© B) = P(5, 5) + P(6, 6) = P(5).P(5) + P(6).P(6)

= 0.1 x 0.1 + 0.1 x 0.1 = 0.02

From eq. (i) we get

0.0 2 = 0. 20 x 0 .1 0

0.02 = 0.02

Hence, A and B are independent events.

Ans.

âˆ´ n(A) = 6 and n{S) = 6

âˆ´

and B = {(4.6).(6,4),(5.5).(6.5).(5,6),(6.6)}

â‡’ n(B) = 6

âˆ´

Also, (A âˆ© B) = {(5, 5), (6,6)}

âˆ´

Also,

Thus, P(A âˆ© B) â‰ P(A).P(B)

So. A and B are not independent events.

Q.3. The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate.

Ans.

P(A âˆª B ) = 0.6 and P{A âˆ© B) = 0.3

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

Ans.

For atleast one of the three marbles drawn be black, if the first marble is red, then the following situations are possible

Event E

Event E

Event E

Let Event R

Event B

Ans.

âˆ´ n(S) = 36, where S is the sample space.

Event â€˜E' is â€˜a total o f 4 â€™

E = {( 2, 2), ( 3, 1), ( 1, 3)}x

Event â€˜F' is la total of 9 or moreâ€™

âˆ´ F= {(3,6), (6,3), (4, 5), (5,4). (4, 6). (6,4), (5, 5), (5,6), (6, 5), (6,6)}

Event 'G' is â€˜a total divisible by 5â€™

âˆ´ G - {(1,4), (4, 1), (2,3), (3,2), (4,6), (6,4), (5, 5)}

Here, (E âˆ© F) = Ï† and (E âˆ© G) = Ï†

Also, ( F âˆ© G ) = {(4, 6), ( 6 .4 ). (5, 5)}

and

So,

Hence, there is no pair which is independent.

Ans.

(i) There are 2 outcomes for each trial

(ii) There is a fixed number of trials

(iii) The probability of success must be the same for all the trials.

When coin is tossed, possible outcomes are Head and Tail. Since coin is tossed three times, we have fixed number of trials. Also probability of Head and Tail in each trial is 1/2. Thus given experiment is said to have binomial distribution.

Find:

(1) P(A|B)

(2) P(B|A)

(3) P(A'|B)

(4) P(A'|B')

Ans.

Ans.

(1) p

(2) (1â€“p

(3) 1â€“(1â€“p

(4) p

Ans.

So, E

So, E

So, either E

So, either E

(1) Find the value of k

(2) Determine the mean of the distribution.

â‡’ k + k

â‡’ 3k

â‡’ 3k

â‡’ 3k(k + 1) - 1(k + 1) = 0

â‡’ (3k - 1)(k + 1) = 0

âˆ´ k = 1/3 and k = â€“ 1

But k â‰¥ 0

âˆ´ k = 1/3

R.H.S. = P(A âˆ© B) +

= P(A).P(B) + P(A) .= P(A) [P(B) +]

= P(A) . 1

= P(A) = L.H.S. Hence proved.

R.H.S. = P(A) . P(B) + P(A) .+ . P(B)

= P(A).P(B) + P(A)[1 â€“ P(B)] + [1 â€“ P(A)].P(B)

= P(A).P(B) + P(A) â€“ P(A).P(B) + P(B) â€“ P(A).P(B)

= P(A) + P(B) â€“ P(A âˆ© B)

= P(A âˆª B) = L.H.S. Hence proved.

Probability distribution table is:

we know that Var(X) =

âˆ´ Standard deviation =

Since, she loses Rs.1 for getting any of 2, 4, 5.

So,

Player â€™s expected profit =

âˆ´ Sample space n(S) = (6)

Let E

â‡’ E

â‡’ n(E

Hence, expectation is Rs.0.65.

And E be the event that the ball drawn from the second bag is white.

Hence, the required probability is 5/9.

and bag II = {2 B, 4 W}

Let E

E

and E = The event that a black ball is selected

Hence, the required probability is 7/15.

Let E

E

and E is the event that second ball drawn is blue.

âˆ´ P(E) = P(E

Hence, the required probability is 5/9.

âˆ´ P(E

Hence, the required probability is 1/27075.

and n = 5

Hence, the required probability is 5/16.

Hence, the required probability is 7/128.

Hence, the required probability is 4547/8192.

Here, n = 8, p = 1/10 and q = 1 - 1/10 = 9/10 and r â‰¥ 1

P(X â‰¥ 1) = 1 - P(x = 0) =

Hence, the required probability is .

Calculate (1)

(2) Variance of X.

We know that: Var(X) = E(X

where E(X) = and E(X

âˆ´ E(X) = 0 Ã— 0.1 + 1 Ã— 0.25 + 2 Ã— 0.3 + 3 Ã— 0.2 + 4 Ã— 0.15

= 0 + 0.25 + 0.6 + 0.6 + 0.6 = 2.05

E(X

= 0 + 0.25 + 1.2 + 1.8 + 2.40 = 5.65

(ii) Var(X) = 1.4475

(1) Determine the value of k.

(2) Determine P(X â‰¤ 2) and P(X > 2)

(3) Find P(X â‰¤ 2) + P (X > 2).

âˆ´ k = 8/15

(2)

(3)

We know that V(X) = E(X

âˆ´

Hence, the required variance = 0.18.**Q.28. A die is thrown three times. Let X be â€˜the number of twos seenâ€™. Find the expectation of X.****Ans. **Here, we have X = 0, 1, 2, 3 [âˆ´ die is thrown 3 times]

and p = 1/6, q = 5/6

âˆ´P(X = 0) = P(not 2) . P(not 2) . P(not 2) =

P(X = 1) = P(2) . P(not 2) . P(not 2) + P(not 2) . P(2) . P(not 2) + P(not 2) . P(not 2) . P(2)

Now

Hence, the required expectation is 1/2.**Q.29. Two biased dice are thrown together. For the first die P(6) = 1/2, the other scores ****being equally likely while for the second die, P(1) = 2/5 and the other scores are equally likely. Find the probability distribution of â€˜the number of ones seenâ€™.****Ans. **Given that: for the first die, P(6) = 1/2 and= 1- 1/2 = 1/2

For the second die, P(1) = 2/5 and = 1- 2 =3/5

Let X be the number of oneâ€™s seen

âˆ´ X = 0, 1, 2

Hence, the required probability distribution is**Q.30. Two probability distributions of the discrete random variable X and Y are given below.Prove that E(Y**

We know that, E(X) =

For the second probability distribution,

Now E(Y^{2}) = 14/5 and 2 E(X) = 2. 7/5 = 14/5

Hence, E(Y^{2}) = 2E(X).**Q.31. A factory produces bulbs. The probability that any one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability that(i) None of the bulbs is defective(ii) Exactly two bulbs are defective(iii) More than 8 bulbs work properly**

Here, n = 10,

We know that P(X = r) =

(i) None of the bulbs is defective, i.e., r = 0

(ii) Exactly two bulbs are defective

(iii) More than 8 bulbs work properly We can say that less than 2 bulbs are defective

P(x < 2) = P(x = 0) + P(x = 1)

Let E

E

and H = Event that the tossed coin gets head.

âˆ´ Using Bayesâ€™ Theorem, we get

Hence the required probability is 1/3.

E

and H = The event that selected person is left handed

âˆ´ P(E

P (HE

So, from Bayesâ€™ Theorem

Hence, the required probability is 9/44.

âˆ´

Hence, the required probability is

X = 1, 2, 3, 4, 5, 6 and S = {(1, 1), (1, 2), (2, 1), (2, 2), (1, 3), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), ..., (6, 6)}

So,

Similarly

So, the required distribution is

Now, the mean E(X) =

Hence, the required mean = 161/36.

and P(X) at X = 0 and 1 is p. Let P(X) at X = 2 is x

â‡’ p + p + x = 1

â‡’ x = 1 â€“ 2p

Now we have the following distributions.

âˆ´ E(X) = 0.p + 1.p + 2(1 â€“ 2p) = p + 2 â€“ 4p = 2 â€“ 3p

and E(X

Given that: E(X

âˆ´ 4 â€“ 7p = 2 â€“ 3p

â‡’ 4p = 2

p = 1/2

Hence, the required value of p is 1/2.

Variance (X) = E(X

Hence, the required variance is 665/324.

= {(2, 4), (4, 2), (1, 5), (5, 1), (3, 3)}

and B

= {(2, 5), (5, 2), (1, 6), (6, 1), (3, 4), (4, 3)}

Let P(A

and P(B

âˆ´ The required probability of winning A in his third throw

where (x, y) denotes a typical sample point.

A = {(x, y) : x + y = 11} and B = {(x, y) : x â‰ 5}

âˆ´ A = {(5, 6), (6, 5)},

B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

â‡’ n(A) = 2, n(B) = 30 and n(A âˆ© B) = 1

âˆ´ P(A) = 2/36 = 1/18 and P(B) = =30/36 = 5/6

â‡’ P(A). P(B) = 1/18.5/6 = 5/108 and P(A âˆ© B) = 1/36

Since P(A). P(B) â‰ P(A âˆ© B)

Hence, A and B are not independent.

E

E

E

âˆ´

Now P(E

Hence, the probability of drawing a white ball does not depend upon k.

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