Question

Maxine plays a game in which the number of points she receives is given by the discrete random variable X whose values are shown in the table below. The probability of each value of X is given by (0.4 – 0.1X) e.g. P(1) = 0.4 – 0.1(1) = 0.3

X	0	1	2	3
P		0.3
X2

1. Copy and complete the Table above.
2. What is the range of the points?
3. What is mode of the points?
4. Find E[X], E[X²] and Var[X], the Variance of X.
5. Find Var[5 – 2X] .

Maxine can win a prize if the total number of points she has scored after 5 games is at least 10. After 3 games, which are independent, Maxine has a total of 6 points.

(vi) Find the probability that Maxine will win a prize.

(b)     Suppose on a particular day the market will independently either Rise, R with probability 0.55, or Fall, F with probability 0.45. Consider a sequence of 4 consecutive days.

1. List all the sequences in which the market rises on only 1 of the 4 days.
2. What is the probability of your first listed sequence occurring?
3. What is the probability of any one of your listed sequences occurring?
4. Explain why your answer to (iii) is unchanged if four random days are selected instead of four consecutive days.

Answer 1

Maxine plays a game in which the number of points she receives is given by the discrete random variable X.

Also, $P(X)=(0.4-0.1*X)$

(i)

X	0	1	2	3
P(X)	0.4	0.3	0.2	0.1
X2	0	1	4	9

(ii)

Range of the points:

Range = 3-0 = 3

(iii)

Mode i.e. the most probable point is $X=0$

(iv)

3 Ε[X] - Στ και Ρ(x) T=0

=0+ (1+0.3)+(2*0.2)+(3+0.1] = 1.0

$E[X^2]=\sum_{x=0}^{3}x^2*P(x)$

$=0+(1*0.3)+(2^2*0.2)+(3^2*0.1)=2.0$

$Var(X)=E(X^2)-E^2(X)$

$=2.0-1.0^2=1.0$

(v)

$Var[5-2X]=0+(-2)^2*Var(X)=4.0$

Also, Maxine can win a prize if the total number of points she has scored after 5 games is at least 10. After 3 games, which are independent, Maxine has a total of 6 points. R with probability 0.55, or Fall, F with probability 0.45. A sequence of 4 consecutive days is considered.

(vi)

Probability that maxine will a prize

=P(4 points in next 2 independent games, when 6 points is earned in previous 3 independent games)

=P(4 points in 2 independent games)

=P[(X =1 in 1st game &  X=3 in 2nd game) OR (X =2 in 1st game &  X=2 in 2nd game)OR(X =3 in 1st game &  X=1 in 2nd game)]

$=P(X_1=1,X_2=3)+P(X_1=2,X_2=2)+P(X_1=3,X_2=1)$

$=P(X=1)*P(X=3)+P(X=2)*P(X=2)+P(X=3)*P(X=1)$

$=0.3*0.1+0.2*0.2+0.1*0.3=0.03+0.04+0.03=0.1$

I.

The sequence where the market rises on only 1 of the 4 days:

II.

Probability of first listed sequence occurring

$=P[RFFF]=$$=P[RFFF]=0.55*0.45*0.45*0.45=0.05011\approx 0.0501$

III.

Probability of any one of listed sequence occurring

$=P[RFFF\:or\:FRFF\:or\:FFRF\:or\:FFFR]$

$=0.0501$

IV.

(iii) is unchanged if four random days are selected instead of four consecutive days because the each of the rise or fall is independent of each other.