Discrete random variables

The probability distribution for the random variable X is given in the following table.

Find the value of p and work out P(2<X<5) 

Sum of probability is

`p + 2p + 1/2p + 3p = 1`

`p = 2/13`

P(2<X<5) = P(X = 3 or 4) = `4/13+1/13 = 5/13`

The probability distribution for the random variable W is given in the following table

a.  Form an equation using k, then solve it.

b.  Explain why only one of your solutions is valid.

c.  Find

a. Sum of probabilities is

`2k+k^2 +k/2 +(4/5-3k) + 13/50 = 1`

`50k^2 - 25k +3 = 0`

`(10k-3)(5k-1) = 0`

b. 

Probabilities cannot be negative, so  is the only valid solution

`P(6 le W < 10) = P(W =6 or 9) = 0.04 + 0.1 = 0.14`

The discrete random variable R is such that `R in {1, 3, 5, 7}`

a. Given that P(R = r) = `frac(k(r+1))(r+2)`, find the value of the constant k

b. Hence, find `P( R le 4)`

a. `P(R = 1,3,5 or 7) = (2k)/3 + (4k)/5 + (6k)/7 + (8k)/9 = frac(1012k)(315) = 1`

b. `P(R le 4) = P(R = 1 or 3) = 105/506 + 126/506 = 21/46`

The probability distribution for the random variable X is given in the following table.

Calculate E(X) and Var(X)

E(X) = `sumxp= 0*0.1 + 1*0.12 + 2*0.36 + 3*0.42= 2.1`

Var(X) = `sumx^2p- {E(X)}^2`

`= 0^2*0.1 + 1^2*0.12 + 2^2*0.36 = 3^2*0.42 - 2.1^2 = 0.93`

The probability distribution for the random variable W is given in the following table.

Given that E(W) = a, find a and evaluate Var(W)

E(W) = `0.6*2.1 + 0.1*a + 7.2 = a`, which gives a = 11

Var(W) = `10^2*0.05 + 20^2*0.1 +70^2*0.35 +100^2*0.5 - 77^2 = 831`

A company offers a $1000 cash loan to anyone earning a monthly salary of at least $2000. To secure the loan, the borrower signs a contract with a promise to repay the $1000 plus a fixed fee before 3 months have elapsed. Failure to do this gives the company a legal right to take $1540 from the borrower’s next salary before return ing any amount that has been repaid.

From past experience, the company predicts that 70% of borrowers succeed in repaying the loan plus the fixed fee before 3 months have elapsed.

a.  Calculate the fixed fee that ensures the company an expected 40% profit from each $1000 loan.

b.  Assuming that the company charges the fee found in part a, how would it be possible, without changing the loan conditions, for the company’s expected profit from each $1000 loan to be greater than 40%? 

a. Expected profit is made from a combination of fixed fees ($x) and claims of $540

`0.7x + (0.3*540) = 400`gives x = fixed fee = $340 

b. If the successful repayment rate is , profit is

240r + 540(1-r) > 400 

-200r > -140 

r < 0.7 = 70%

E(profit) > 40% if successful repayment rate is below 70%

An ordinary fair die is rolled. If the die shows an odd number then S, the score awarded, is equal to that number. If the die shows an even number, then the die is rolled again. If on the second roll it shows an odd number, then that is the score awarded. If the die shows an even number on the second roll, the score awarded is equal to half of that even number.

a  List the possible values of S and draw up a probability distribution table.

b  Find P[S>E(S)].

c  Calculate the exact value of Var(S).

a. `S in {1,2,3,5}`

P(S = 1) = `1/6 + (3/6*2/6) = 1/3`

P(S = 2) = `3/6*1/6 = 1/12`

P(S = 3) = `1/6 + (3/6*2/6) = 1/3`

P(S = 5) = `1/6 + (3/6*1/6) = 1/4`

b. E(S) = `1/12*(4+ 2+12+15) = 2frac3(4) or 2.75`

P[S > E(S)] = P(S = 3 or 5) = `4/12 + 3/12 = 7/12`

c. Var(S) = `1/12*(1^2 *4 + 2^2*1 + 3^2*4 + 5^2*3) - (2frac3(4))^2`

`= 2frac(17)(48) or 113/48`

Find the mean and the variance of the discrete random variable X , whose probability distribution is given in the following table.

Sum of probabilities is

`1-k + 2-3k + 3-4k+ 4-6k = 10 -14k =1`

`k = 9/14`

Mean = E(X) = `1*5/14 + 2*1/14+ 3*6/14 + 4*2/14 =2frac(5)(14) or 2.36`

Var(X) = `1^2 *5/14 + 2^2*1/14 +3^2*6/14 + 4^2*2/14 - (2frac(5)(14))^2 = 1frac(45)(196) or 1.23`

The following table shows the probability distribution for the random variable Y

a. Given that Var(Y) = 1385.2, show that `q^2 - 61q + 624 = 0` and solve this equation

b. Find the greatest possible value of E(Y) 

a. E(Y) = `1*0.2 + 10*0.4 + 0.2q + 101*0.2 = 0.2q + 24.4`

Var(Y) = `1^2*0.2 + 10^2*0.4 + 0.2q^2 + 101^2*0.2 -(0.2q+24.4)^2`

Var(Y) = `0.16q^2 - 9.76q + 1485.04`

Then we have: 

`0.16q^2 - 9.76q + 1485.04 = 1385.2`

`q^2 - 61q + 624 = 0`

(q-13)(q-48) = 0 

q = 13 or q = 48 

b. q = 13 gives E(Y) = `0.2*13 + 24.4 = 27`

q = 48 gives E(Y) = `0.2*48 + 24.4 = 34`

Therefore, the greatest possible value of E(Y) is 34. 

An investment company has produced the following table, which shows the probabilities of various percentage profits on money invested over a period of 3 years

a.  Calculate the expected profit on an investment of $50 000.

b.  A woman considers investing $50 000 with the company, but decides that her money is likely to earn more when invested over the same period in a savings account that pays r% compound interest per annum. Calculate, correct to 2 decimal places, the least possible value of r.

a. E(%profit) = `sum(%*p) = 0.05 + 0.5 + 5 + 3 + 1 + 1.2 + 0.9 + 0.5 = 13.35%`

E(profit) = `0.1335*50000 = $6675`

b. `(1 + r/100)^3 *50000 > 56675`

1 + `r/100 > root(3)(frac(56675)(50000))`

r > 100`*(root(3)(1.1335) - 1) = 4.2654`, so least possible r is 4.27. 

A chef wishes to decorate each of four cupcakes with one randomly selected sweet. They choose the sweets at random from eight toffees, three chocolates and one jelly.

Find the variance of the number of cupcakes that will be decorated with a chocolate sweet.

Let the number of cakes decorated with a chocolate sweet by

P(X = 0) = `frac(C_3^0*C_9^4)(C_12^4) = 126/495`

P(X = 1) = `frac(C_3^1*C_9^3)(C_12^4) = 252/495`

P(X = 2) = `frac(C_3^2*C_9^2)(C_12^4) = 108/495`

P(X = 3) = `frac(C_3^3*C_9^1)(C_12^4) = 9/495`

E(X) = `0*126/495 + 1*252/495 + 2*108/495 + 3*9/495 = 1`

Var(X) = `0^2*126/495 + 1^2*252/495+2^2*108/495 + 3^2*9/495 -1^2 =6/11`

The faces of a biased dice are numbered 1, 2, 3, 4, 5 and 6. The random variable X is the score when the die is thrown. The probability distribution table for X is given.

The dice is thrown 3 times. Find the probability that the score is at least 4 on at least 1 of the 3 throws.

Sum of probability is 4p + 0.4 = 1 , which gives p = 0.15

`P(X ge 4) = 0.15 + 0.2 + 0.2 = 0. 55`

P(`X ge 4` on at least one three) = 1 - P( X < 4 on all three) 

= 1 - `(1-0.55)^3 = 0.909`

A picnic basket contains five jars: one of marmalade, two of peanut butter and two of jam. A boy removes one jar at random from the basket and then his sister takes two jars, both selected at random.

a.  Find the probability that the sister selects her jars from a basket that contains:

i.  exactly one jar of jam

ii.  exactly two jars of jam

b.  Draw up the probability distribution table for J, the number of jars of jam selected by the sister, and show that  

a.i. P(contains 1 jam) = P(boy removes jam) = `2/5`

ii. P(contains 2 jam) = P(boy removes marmalade or peanut butter) = `3/5`

b. P( J = 0) = `2/5*(3/4*2/3) + 3/4*(2/4*1/3) = 0.3`

P(J=1) = `2/5*(3/4*1/3+1/4*3/3) + 3/5*(2/4*2/3+2/4*2/3) = 0.6`

P(J =2) = `3/5*(2/4*1/3) = 0.1`

E(J) = `0*0.3 + 1*0.6 + 2*0.1`= 0.8 

Two ordinary fair dice are rolled. The product and the sum of the two numbers obtained are calculated. The score awarded, S, is equal to the absolute (i.e. non-negative) difference between the product and the sum. For example, if 5 and 3 are rolled, then S = `5*3 - (5+3) = 7`

a.  State the value of S when 1 and 4 are rolled

b.  Draw up a table showing the probability distribution for the 14 possible values of S, and use it to calculate E(S)

a. S = `(4+1) - (4*1) = 1`

b. 

E(S) = `frac(0 + 13+4 +9 + 8 + 10 + 14 + 8 + 18 + 22 + 28 + 15 + 38 + 24)(36`

= `5frac(31)(36) or 5.86`

A fair triangular spinner has sides labelled 0, 1 and 2, and another fair triangular spinner has sides labelled –1, 0 and 1. The score, X , is equal to the sum of the squares of the two numbers on which the spinners come to rest.

a.  List the five possible values of X

b.  Draw up the probability distribution table for X

c.  Given that , find the probability that a score of 1 is obtained with at least one of the spinners

d. Find the exact value of a, such that the standard deviation of X is `1/a *E(X)`

a. 0, 1, 2, 4, 5

b. The grid shows values of X

c. There are six ways to score < 4, and at least one spinner scores 1 in four of these.

P(X < 4) = `4/6 = 2/3`

d. E(X) = `1/9 *(0 + 3+ 4+4+ 10) = 7/3`

SD(X) = `sqrt(1/9*(0^2 + 3^2+4^2 + 4^2+10^2)-(7/3)^2) = sqrt(28/9) = frac(2sqrt7)(3`

`frac(2sqrt7)(3) = 1/a*7/3`gives a = `sqrt7/2`

A discrete random variable X, where `X in {2, 3, 4, 5}`, is such that P(X = x) = `frac((b-x)^2)30`

a. Calculate the two possible values of

b. Hence, find P(2<X<5)

a. Sum of probability is

`frac((b-2)^2)(30) + frac((b-3)^2)(30) + frac((b-4)^2)(30) + frac((b-5)^2)(30) = 1`

`(b-2)^2 + (b-3)^2 (b-4)^2 (b-5)^2 = 30`

`4b^2 - 28b + 24 = 0`

4(b-1)(b-6) = 0 

b = 1 or b = 6 

b. P(2<X<5) = P(X =3 or 4) = `9/30+4/30 = 13/30`

X is a discrete random variable and `X in {0, 1, 2, 3}`

Given that P(X > 1) = 0.24, P(0<X<3) = 0.5 and P(X = 0 or 2) = 0.62. 

Find `P(X le 2|X>0).`

Let a, b, c  and d represent P(X=0), P(X = 1), P(X = 2) and P(X = 3) respectively

We have 

a + b+ c + d = 1

c + d = 0.24 

b + c = 0.5 

a + c = 0.62 

Solving these four equations gives a = 0.44, b = 0.32, c = 0.18 and d = 0.06 

P(X>0) = `frac(P(X>0 and X le 2))(P(X>0)) = frac(P(X = 1 or 2))(P(X = 1, 2 or 3)) = frac(0.32 + 0.18)(0.32 + 0.18+ 0.06) = 25/28 or 0.893`

Four students are to be selected at random from a group that consists of seven boys and x girls. The variables B and G are, respectively, the number of boys selected and the number of girls selected.

a Given that P(B = 1) = P(B = 2) , find the value of x

b Given that `G ne 3`, find the probability that G = 4

a. `P(B =1 ) = 2*7/(x+7)*x/(x+6) * (x-1)/(x+5) * (x-2)(x+4)`

`= frac(28x(x-1)(x-2))((x+7)(x+6)(x+5)(x+4))`

Since P(B=1) = P(B=2). Therefore 

`frac(28x(x-1)(x-2))((x+7)(x+6)(x+5)(x+4)) = frac(252x(x-1))((x+7)(x+6)(x+5)(x+4)`

`28(x-2) = 252`

x = 11

b. Select four from 18 students (11 girls and seven boys)

`P(G ne 3) = 1 - P(G=3) =1 - frac(C_11^3*C_7^1)(C_18^4) = 127/204`

P(G = 4) = `frac(C_11^4*C_7^0)(C_18^4) = 11/102`

`P(G ne 3) = frac(P(G ne 3 and G = 4))(P(G ne 3)) = frac(P(G = 4))(P (G ne 3))= 11/102div127/204 = 22/127`

A box contains 2 green apples and 2 red apples. Apples are taken from the box, one at a time, without replacement. When both red apples have been taken, the process stops. The random variable X is the number of apples which have been taken when the process stops. 

i.  Show that `P(X = 3) = 1/3`

ii. Draw up the probability distribution table for

Another box contains 2 yellow peppers and 5 orange peppers. Three peppers are taken from the box without replacement.

iii. Given that at least 2 of the peppers taken from the box are orange, find the probability that all 3 peppers are orange. 

i. P(X = 3) = P(RGR) + P(GRR) = `(2/4*2/3*1/2)(2/4*2/3*1/2) = 1/3`

ii. 

P(X = 2) = P(RR) = `2/4*1/3= 1/6`

P(X = 4) = P(GGRR) + P(GRGR) + P(RGGR) = `3 *(2/4*1/3*2/2*1/1) = 1/2`

iii. Let the number of orange peppers selected be J

`P( J ge 2) = P(J = 2 or 3) =``frac(C_5^2*C_2^1)(C_7^3) + frac(C_5^3*C_2^0)(C_7^3) = 4/7 + 2/7 = 6/7`

`P(J ge 2) = frac(P(J ge 2 and J = 3))(P( J ge 2)) = frac(P(J =3))(P(J ge2)) = 2/7 div 6/7 = 1/3`

In a particular discrete probability distribution, the random variable  takes the value `120/r` with probability `r/45`, where r takes all integer values from 1 to 9 inclusive 

i. Show that P(X = 40) = `1/15`

ii.  Construct the probability distribution table for

iii.  Which is the modal value of X ?

iv.  Find the probability that X lies between 18 and 100

i. `120/r = 40,` so r = 3; P(X = 40) = `3/45 = 1/15`

ii. P(X = x) = `120/(45x) = 8/(3x)`

iii. The modal value has the highest relative frequency (probability), which is `13frac1 3`

iv. P(18<X<100) = `6/45 + 5/45 + 4/45 + 3/45 + 2/45 = 4/9 or 0.444`