1

The lifetime of two brands of bulbs $X$ and $Y$ are exponentially distributed with the mean life of $100$ hours. Bulb $X$ is switched on $15$ hours after Bulb $Y$ has been switched on. The probability that bulb $X$ fails before Bulb $Y$ is?

The options are:

$(A). \frac{15}{100}$

$(B). \frac{1}{2}$

$(C).\frac{85}{100}$

$(D). 0$

Harsh Kumar
  • 2,924
Bridget
  • 101

2 Answers2

4

If bulb $Y$ might fail before bulb $X$ is switched on,

\begin{align} &\text{ probability that bulb $X$ fails before bulb } Y \\ &= P(Y \geq 15) P(X<Y-15|Y \geq 15)+P(Y < 15)(0) \\&=P(Y \geq 15) P(X < Y) \\ &= \frac12 P(Y \geq 15) \end{align}

If we are given that bulb $Y$ is still working when bulb $X$ is switched on \begin{align} &\text{ probability that bulb $X$ fails before bulb } Y \text{ given that bulb } Y \text{ was functioning when bulb } X \text{was switched on} \\ &= P(X<Y-15|Y \geq 15) \\&= P(X < Y) \\&= \frac12 \end{align}

Siong Thye Goh
  • 153,832
  • 2
    Actually the options (rational and including $\frac12$) indicate that the condition $Y\geq15$ is abusively not mentioned. Secondly in the title the OP asks for $P(X<Y)$. I think that must be $P(X<Y-15\mid Y\geq15)$ (as you remark). – drhab Jan 27 '17 at 09:54
  • 1
    +1, interpreting the question as asking for $P(Y>X+15 | Y>15)$ is the only way I can see to make it fit the given answers. – Matthew Towers Jan 27 '17 at 15:51
  • I still don't understand why $P(X<Y)=\frac12$ Is this a $P(X<Y)=\frac{\frac{1}{100}}{\frac{1}{100}+\frac{1}{100}}$ situation? – Diya Sep 28 '19 at 09:11
  • They are two continuous IID, hence the value is $\frac12$. – Siong Thye Goh Sep 28 '19 at 10:11
1

Edit: (Bulb X is switched on 15 hours after bulb Y)

You must partition on the 15 hour.   Bulb-X may fail before bulb-Y, only if bulb-Y does not fail before bulb X is switched on. $$\mathsf P(X<Y)~=~\mathsf P(Y\geq 15)\cdot\mathsf P(X<Y\mid Y\geq 15)$$ Now, use the memoryless property to simplify.

Graham Kemp
  • 133,231
  • Concerning LHS: since $X,Y$ are iid and continuous, it is immediate that $P(X<Y)=\frac12$. Concerning RHS: it seems you accidently switched $X$ and $Y$. – drhab Jan 27 '17 at 10:10
  • $X$ and $Y$ are not iid because bulb $X$ is switched on 15 hours later. $Y$ and $X-15$ are iid. (But yes, I did have the switch on times mixed around) – Graham Kemp Jan 27 '17 at 13:46