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Ex 8A

7. Sam's pocket contains one £1 coin, one 50p coin and three 20p coins. He selects 2 coins at random to place in a collection box. The random varibale X represents the amount, in pence, that he puts in the box.

a) Show that P(X=70) = 0.3

b) Find the probablity distribution for X.

8) A fair coin is tossed repeatedly until a head appears or 3 tosses have been made. The random variable T represents the no. of tosses.

a) Show that P(X=2) = 0.25

b) Find the probablity distribution for T.

The random variable H represents the no. of heads.

c) Find the probablity distribution for H.

11. The discrete random variable X has c.d.f. F (x) defined by

F (x)= 2x+1 ; x=0,1,2,3 and 4

..........7

a) Find F(3)

b) Show that P(X=2) = 2/9

c) Find the probablity distribution for X.

14. A darts player practises throwing a dart at the bull's eye on a dart board. Independently for each throw, her prob. of hitting the bull' s eye is 0.2. Let X be the no. of throws she makes, up to and including her first success.

a) Find the probablity that she is successful for the first time on her 3rd throw.

b) Write down an equation for the prob. distribution of X.

c) Find the probablity that she will have at least 3 failures before her first success.

15. 6 fuses, of which 2 are defective and 4 are good, are to be t5ested one after another in random order until both defective fuses are identified. Find the probablity that the no. of fuses that will be tested is:

a) three

b) four or fewer

Ex 8B

5b) A fair die is rolled and the random variable N represents the no. showing. A square of side N is then drawn on a piece of paper.

Find the variance of the perimeter of this square.

PLS HELP. gosh im so crap...

7. Sam's pocket contains one £1 coin, one 50p coin and three 20p coins. He selects 2 coins at random to place in a collection box. The random varibale X represents the amount, in pence, that he puts in the box.

a) Show that P(X=70) = 0.3

b) Find the probablity distribution for X.

8) A fair coin is tossed repeatedly until a head appears or 3 tosses have been made. The random variable T represents the no. of tosses.

a) Show that P(X=2) = 0.25

b) Find the probablity distribution for T.

The random variable H represents the no. of heads.

c) Find the probablity distribution for H.

11. The discrete random variable X has c.d.f. F (x) defined by

F (x)= 2x+1 ; x=0,1,2,3 and 4

..........7

a) Find F(3)

b) Show that P(X=2) = 2/9

c) Find the probablity distribution for X.

14. A darts player practises throwing a dart at the bull's eye on a dart board. Independently for each throw, her prob. of hitting the bull' s eye is 0.2. Let X be the no. of throws she makes, up to and including her first success.

a) Find the probablity that she is successful for the first time on her 3rd throw.

b) Write down an equation for the prob. distribution of X.

c) Find the probablity that she will have at least 3 failures before her first success.

15. 6 fuses, of which 2 are defective and 4 are good, are to be t5ested one after another in random order until both defective fuses are identified. Find the probablity that the no. of fuses that will be tested is:

a) three

b) four or fewer

Ex 8B

5b) A fair die is rolled and the random variable N represents the no. showing. A square of side N is then drawn on a piece of paper.

Find the variance of the perimeter of this square.

PLS HELP. gosh im so crap...

(7)

Imagine that the 20p coins are distinguished from one other - call them "20p coin A", "20p coin B" and "20p coin C". Then there are (5 choose 2) = 5!/(2!3!) = 10 ways of choosing two coins.

20p(A) + 20p(B) -------> X = 40

20p(A) + 20p(C) -------> X = 40

20p(A) + 50p -------> X = 70

20p(A) + £1 -------> X = 120

20p(B) + 20p(C) -------> X = 40

20p(B) + 50p -------> X = 70

20p(B) + £1 -------> X = 120

20p(C) + 50p -------> X = 70

20p(C) + £1 -------> X = 120

50p + £1 -------> X = 150

These 10 possibilities are equally likely. The probability distribution for X is

P(X = 40) = (Number of possibilities where X = 40)/10 = 3/10

P(X = 70) = (Number of possibilities where X = 70)/10 = 3/10

P(X = 120) = (Number of possibilities where X = 120)/10 = 3/10

P(X = 150) = (Number of possibilities where X = 150)/10 = 1/10

(8)

(a)

P(T = 2) = P(first toss tails, second toss heads) = (1/2)*(1/2) = 0.25

(b)

We know P(T = 2), so we just need P(T = 1) and P(T = 3).

P(T = 1) = P(first toss heads) = 0.5

P(T = 3) = 1 - P(T = 1) - P(T = 2) = 0.25

(c)

P(H = 0) = P(all three tosses are tails) = (1/2)^3 = 1/8

P(H = 1) = 1 - P(H = 0) = 7/8 . . . since there can't be more than one head during the game

(11)

The question should say "9" instead of "7".

(a) F(3) = (2*3 + 1)/9 = 7/9

(b) P(X = 2) = P(X <= 2) - P(X <= 1) = F(2) - F(1) = 5/9 - 3/9 = 2/9

(c)

P(X = 0) = P(X <= 0) = F(0) = 1/9

P(X = 1) = P(X <= 1) - P(X <= 0) = 3/9 - 1/9 = 2/9

P(X = 2) = P(X <= 2) - P(X <= 1) = 5/9 - 3/9 = 2/9

P(X = 3) = P(X <= 3) - P(X <= 2) = 7/9 - 5/9 = 2/9

P(X = 4) = P(X <= 4) - P(X <= 3) = 1 - 7/9 = 2/9

(14)

(a)

P(X = 3) = P(first two throws miss, third throw hits) = 0.8*0.8*0.2 = 0.128

(b)

X has a geometric distribution.

P(X = x) = P(first x - 1 throws miss, xth throw hits) = 0.8^(x - 1) * 0.2

(c)

P(she has at least 3 failures before her first success)

= P(first three throws miss)

= 0.8^3

= 0.512

(15)

(a)

P(first three tests are DGD) = (2/6)(4/5)(1/4) = 1/15

P(first three tests are GDD) = (4/6)(2/5)(1/4) = 1/15

Answer = 1/15 + 1/15 = 2/15

(b)

A testing order can be represented by a permutation of the six letters DDGGGG. There are 6!/(2!4!) = 15 such orders. These 15 possibilities are equally likely.

Four or fewer tests will be required

if and only if

the permutation ends in GG.

There are 4!/(2!2!) = 6 permutations that end in GG (= the number of permutations of DDGG).

Answer = 6/15 = 2/5.

(5b)

We need two formulas:

Var(X) = E(X^2) - [E(X)]^2

Var(aX) = a^2 Var(X)

Those formulas work for any random variable X and any number a.

E(N)

= (1 + 2 + 3 + 4 + 5 + 6)/6

= 7/2

E(N^2)

= (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2)/6

= 91/6

Var(N)

= E(N^2) - [E(N)]^2

= 35/12

Let P = 4N = perimeter of the square. Then

Var(P)

= Var(4N)

= 4^2 Var(N)

= 16 Var(N)

= 16 (35/12)

= 140/3

Imagine that the 20p coins are distinguished from one other - call them "20p coin A", "20p coin B" and "20p coin C". Then there are (5 choose 2) = 5!/(2!3!) = 10 ways of choosing two coins.

20p(A) + 20p(B) -------> X = 40

20p(A) + 20p(C) -------> X = 40

20p(A) + 50p -------> X = 70

20p(A) + £1 -------> X = 120

20p(B) + 20p(C) -------> X = 40

20p(B) + 50p -------> X = 70

20p(B) + £1 -------> X = 120

20p(C) + 50p -------> X = 70

20p(C) + £1 -------> X = 120

50p + £1 -------> X = 150

These 10 possibilities are equally likely. The probability distribution for X is

P(X = 40) = (Number of possibilities where X = 40)/10 = 3/10

P(X = 70) = (Number of possibilities where X = 70)/10 = 3/10

P(X = 120) = (Number of possibilities where X = 120)/10 = 3/10

P(X = 150) = (Number of possibilities where X = 150)/10 = 1/10

(8)

(a)

P(T = 2) = P(first toss tails, second toss heads) = (1/2)*(1/2) = 0.25

(b)

We know P(T = 2), so we just need P(T = 1) and P(T = 3).

P(T = 1) = P(first toss heads) = 0.5

P(T = 3) = 1 - P(T = 1) - P(T = 2) = 0.25

(c)

P(H = 0) = P(all three tosses are tails) = (1/2)^3 = 1/8

P(H = 1) = 1 - P(H = 0) = 7/8 . . . since there can't be more than one head during the game

(11)

The question should say "9" instead of "7".

(a) F(3) = (2*3 + 1)/9 = 7/9

(b) P(X = 2) = P(X <= 2) - P(X <= 1) = F(2) - F(1) = 5/9 - 3/9 = 2/9

(c)

P(X = 0) = P(X <= 0) = F(0) = 1/9

P(X = 1) = P(X <= 1) - P(X <= 0) = 3/9 - 1/9 = 2/9

P(X = 2) = P(X <= 2) - P(X <= 1) = 5/9 - 3/9 = 2/9

P(X = 3) = P(X <= 3) - P(X <= 2) = 7/9 - 5/9 = 2/9

P(X = 4) = P(X <= 4) - P(X <= 3) = 1 - 7/9 = 2/9

(14)

(a)

P(X = 3) = P(first two throws miss, third throw hits) = 0.8*0.8*0.2 = 0.128

(b)

X has a geometric distribution.

P(X = x) = P(first x - 1 throws miss, xth throw hits) = 0.8^(x - 1) * 0.2

(c)

P(she has at least 3 failures before her first success)

= P(first three throws miss)

= 0.8^3

= 0.512

(15)

(a)

P(first three tests are DGD) = (2/6)(4/5)(1/4) = 1/15

P(first three tests are GDD) = (4/6)(2/5)(1/4) = 1/15

Answer = 1/15 + 1/15 = 2/15

(b)

A testing order can be represented by a permutation of the six letters DDGGGG. There are 6!/(2!4!) = 15 such orders. These 15 possibilities are equally likely.

Four or fewer tests will be required

if and only if

the permutation ends in GG.

There are 4!/(2!2!) = 6 permutations that end in GG (= the number of permutations of DDGG).

Answer = 6/15 = 2/5.

(5b)

We need two formulas:

Var(X) = E(X^2) - [E(X)]^2

Var(aX) = a^2 Var(X)

Those formulas work for any random variable X and any number a.

E(N)

= (1 + 2 + 3 + 4 + 5 + 6)/6

= 7/2

E(N^2)

= (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2)/6

= 91/6

Var(N)

= E(N^2) - [E(N)]^2

= 35/12

Let P = 4N = perimeter of the square. Then

Var(P)

= Var(4N)

= 4^2 Var(N)

= 16 Var(N)

= 16 (35/12)

= 140/3

sorry wot does this ^ means?

u r really good

ive got some more questions which im stuck but err have u got the S1 Heinemann book? i sorta can't be bothered to type them out

u r really good

ive got some more questions which im stuck but err have u got the S1 Heinemann book? i sorta can't be bothered to type them out

for question 11

is it bcuz they ask for F(3) and not P(3), so u dun need to minus the probability of 2 and 1?

is it bcuz they ask for F(3) and not P(3), so u dun need to minus the probability of 2 and 1?

Jonny W

No I don't have it.

oh thats ok. u mind if i keep askin u questions?

and err can i ask, will variance ever be negative?

I'm doin the normal distributution and i aint sure how to find mean and variance provided with Z and the and X. im so stuck with those negative and positive signs

lost_in_cyberia

oh thats ok. u mind if i keep askin u questions?

and err can i ask, will variance ever be negative?

I'm doin the normal distributution and i aint sure how to find mean and variance provided with Z and the and X. im so stuck with those negative and positive signs

and err can i ask, will variance ever be negative?

I'm doin the normal distributution and i aint sure how to find mean and variance provided with Z and the and X. im so stuck with those negative and positive signs

Feel free to keep asking questions.

Variance is never negative (because it is the expectation of (X - E(X))^2, and that thing is at least 0). If you get a negative answer, you've made a mistake.

I'm not sure about your second paragraph. Z usually represents a normal random variable with mean 0 and variance 1 - called a "standard" normal random variable. If X = aZ + b then the mean and variance of X are b and a^2. But that might not be what you meant.

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