Question

A ship is fitted with three engines `E_1`, `E_2`, and `E_3`,. The engines function independently of each other with respective probabilities `1/2`. `1/4`and `1/4` ,and For the ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let `X_1`, `X_2`, and `X_3` denote respectively the events that the engines `E_1`, `E_2` and `E_3`,are functioning. Which of the following is (are) true?
A. `P(X_(1)^(c )//X)=(7)/(8)`
B. P [ Exactly two engines of the ship are functioning X] `=(7)/(8)`
C. `P(X // X_(2))=(5)/(16)`
D. `P(X//X_(1))=(7)/(16)`

Answer 1

Correct Answer - C
We have,
`P(X_(1))=(1)/(2), P(X_(2))=(1)/(4), P(X_(3))=(1)/(4)`
`P(X)=P((X_(1) cap X_(2) cap X_(3)^(c )) cup (X_(1) cap X_(2)^(c ) cap X_(3)) cap (X_(1)^(c ) cap X_(2) cap X_(3)) cup (X_(1) cap X_(2) cap X_(3)))`
`=P(X_(1) cap X_(2) cap X_(3)^(c ))+P(X_(1) cap X_(2)^(c ) cap X_(3))+P(X_(1)^(c ) cap X_(2) cap X_(3))+P(X_(1) cap X_(2) cap X_(3))`
`=P(X_(1))P(X_(2))P(X_(3)^(c ))+P(X_(1))P(X_(2)^(c ))P(X_(3))+P(X_(1)^(c ))P(X_(2))P(X_(3))+P(X_(1))P(X_(2))P(X_(3))`
`=(1)/(2)xx(1)/(4)xx(3)/(4)+(1)/(2)xx(3)/(4)xx(1)/(4)+(1)/(2)xx(1)/(4)+(1)/(2)xx(1)/(4)xx(1)/(4)=(1)/(4)`
Now, `P(X_(1)^(c )//X)=(P(X_(1)^(c ) cap X))/(P(X))=(P(X_(1)^(c ) cap X_(2) cap X_(3))/(P(X))`
`=((1)/(2)xx(1)/(4)xx(1)/(4))/((1)/(4))=(1)/(8)`
P(Exactly two engines of the ship are functioning `//X)`
`=P{(X_(1)^(c )capX_(2)cap X_(3))cap(X_(1) cap X_(2)^(c )capX_(3))cap(X_(1)capX_(2)capX_(3)^(c ))})/(P(X))`
`=(P(X_(1)^(c ) capX_(2)capX_(3))+P(X_(1)capX_(2)^(c )capX_(3))+P(X_(1)capX_(2)capX_(3)^(c ))/(P(X))`
`=((1)/(2)xx(1)/(4)xx(1)/(4)+(1)/(2)xx(3)/(4)xx(1)/(4)+(1)/(2)xx(1)/(4)xx(3)/(4))/((1)/(4))=(7)/(8)`
`P(X//X_(2))=(P(X cap X_(2))/(P(X_(2))))`
`=(P{(X_(1)^(c ) capX_(2)capX_(3))cap(X_(1)capX_(2)capX_(3)^(c ))cap(X_(1)capX_(2)capX_(3))})/(P(X_(2)))`
`=(P(X_(1)^(c ) capX_(2)capX_(3))+P(X_(1)capX_(2)capX_(3)^(c ))+P(X_(1)capX_(2)capX_(3)))/(P(X_(3)))`
`=((1)/(2)xx(1)/(4)xx(1)/(4)+(1)/(2)xx(1)/(4)xx(3)/(4)+(1)/(2)xx(1)/(4)xx(1)/(4))/((1)/(4))=((5)/(32))/((1)/(4))=(5)/(8)`
`P(X//X_(1))=(P(Xcap X_(1)))/(P(X_(1)))`
`=(P{(X_(1)capX_(2)capX_(3))cap(X_(1)capX_(2)capX_(3)^(c ))cap(X_(1)capX_(2)^(c )capX_(3))})/(P(X_(1)))`
`=(P(X_(1)capX_(2)capX_(3))+P(X_(1)capX_(2)cap_X(3)^(c ))+P(X_(1)capX_(2)^(c )capX_(3)))/(P(X_(1)))`
`=((1)/(2)xx(1)/(4)xx(1)/(4)+(1)/(2)xx(1)/(4)xx(3)/(4)+(1)/(2)xx(3)/(4)xx(1)/(4))/((1)/(2))=(7)/(16)`