Question

[12] In an experiment on human behavior, a psychologist asks four men and four women to enter a room and sit at a rectangular table. This table has three seats on each of the longer sides of the table, and one seat at each end of the table. The seats at the end of the table are considered to be the dominant seats. (A diagram may help you to visualize this).

a. If the people choose their seats randomly, determine the probability distribution for the random variable X, where X represents the number of women occupying the end seats. [4]

b. Determine E(X) and Var(X). [5]

c. In 15 independent repetitions of the experiment involving different people each time, calculate the probability that women occupy both end seats 2 or more times. [3]

Answer 1

Answer:

a)

Here there are two dominant or we can say the end seats out of eight seats.

there are four women and four men are there.

so here out of these two, women can occupy either no end seats or one end seat or both the end seats.

So here as it given that X represents the number of women occupying the end seats

so here sample space of X is

S(X) = {0, 1, 2}

so here now we have to find the probability distribution of X.

Now it is clear than if X number of women will occupy the end seats (2-X) number of end seats willbe occupied by the men.

so,

p(X) = ²C_x⁴C_X⁴C_(2-X)/⁸C₂

so here

p(0) = ²C₀⁴C₀⁴C₂/⁸C₂ = (4 * 3)/(8 * 7) = 3/14

p(1) = ²C₁ * ⁴C₁⁴C₁/⁸C₂ = (2 * 4 * 4)/(8 * 7) = 8/14

p(2) = ²C₂ * ⁴C₀⁴C₂/⁸C₂ = (4 * 3)/(8 * 7) = 3/14

so pdf of X is

p(x) = 3/14 ;x = 0

= 4/7 ; x = 1

= 3/14 ; x = 2

b)

E[X] = = 3/14 * 0 + 8/14 * 1 + 3/14 * 2 = 1

VaR[X] = E[X²] - E[X]²

E[X²] = = 3/14 * 0 * 0 + 8/14 * 1 * 1 + 3/14 * 2 * 2 = 10/7

VaR[X] = 10/7 - 1² = 3/7

c)

Here now this experiment is performed 15 times so here n = 15

and probability of having women occupy both seats is fixed and it theoretical probability = p = 3/14

so here if y is the number of times women occupy both seats out of 15 times.

then y follows binomial experiment with n = 15 and p = 3/14

so, we have to find

Pr(y >= 2) = BINOMIAL (y > =2 ; n = 15 ; p = 3/14)

p(y) = ¹⁵C_y (3/14)^y(11/14)^(15-y)

Pr(y > = 2) =1 - P(y = 0) - P(y = 1)

= 1 - ¹⁵C₀ (3/14)⁰(11/14)¹⁵ - ¹⁵C₁ (3/14)¹ (11/14)¹⁴

= 1 - 0.0269 - 0.1098 = 0.8633