Question

In: Statistics and Probability

STAT 13_1: A military company has three divisions. There are 20 soldiers in each division. In...

STAT 13_1:

A military company has three divisions. There are 20 soldiers in each division. In divisions A - 4 female soldiers, in divisions B - 8 female soldiers and in divisions C - 12 female soldiers.
During one week, one of the divisions is chosen, incidentally, to be the division that gives a watchman duty.
Each day, one of the soldiers of the selected division is randomly selected to be the division watchman.
(The same soldier or female soldier may be selected for more than one day)

A. What is the probability that on Sunday the division guard will be a female-soldier?
B. Are the events "The selected division is division B" and "On Sunday the Company Guard is a female-Soldier" are independent events? explain!
C. It is known that on Sunday and Monday, the division guard was a female-soldier. What is the probability that the selected division is C?
D. It is known that on Sunday and Monday the duty guard was a female-soldier. What is the probability that on Tuesday the division duty guard will be a soldier (man)?

Solutions

Expert Solution

(A)

From the given data, the following Table is calculated:

Male Female Total
A 16 4 20
B 12 8 20
C 8 12 20
Total 36 24 60

the probability that on Sunday the division guard will be a female-soldier = P(Female) = 24/60 = 0.4

(B)

The events "The selected division is division B" and "On Sunday the Company Guard is a female-Soldier" are independent events

Explanation:

P( The selected division is division B AND On Sunday the Company Guard is a female-Soldier ) = 8/60= 0.1333

P( The selected division is division B ) = 20/60 = 0.3333

P(On Sunday the Company Guard is a female-Soldier) = 24/60= 0.4

So,

P( The selected division is division B ) X P(On Sunday the Company Guard is a female-Soldier) = 0.3333 X 0.4 = 0.1333

Since P( The selected division is division B ) X P(On Sunday the Company Guard is a female-Soldier) = 0.3333 X 0.4 = 0.1333 = P( The selected division is division B AND On Sunday the Company Guard is a female-Soldier ) = 8/60= 0.1333, the events "The selected division is division B" and "On Sunday the Company Guard is a female-Soldier" are independent events

(C)

P(C/ Female) = P(C AND Female)/ P(Female)

                    = 12/24 = 0.5

So,

the probability that the selected division is C = 0.52 = 0.25

(D)

the probability that on Tuesday the division duty guard will be a soldier (man) = 36/60 = 0.6


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