Question

Suppose that a box contains a large number of tacks and that the probability X that a particular tack will land with its point up when it is tossed varies from tack to tack in accordance with the following p.d.f.:

\(f\left( x \right) = \left\{ \begin{array}{l}2\left( {1 - x} \right)\;\;\;\;\;for\;0 < x < 1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Suppose that a tack is selected at random from the box and that this tack is then tossed three times independently. Determine the probability that the tack will land with its point up on all three tosses.

Short answer

Probability that the tack will land with its point up on all three tosses is \(\frac{1}{{10}}\)

Step-by-step solution

Given information:

X be the probability that a particular tack will land with its point up when it is tossed varies from back to back.

 Calculate the probability that the tack will land with its point up on all three tosses.

Let given

\(f\left( x \right) = \left\{ \begin{array}{l}2\left( {1 - x} \right)\;\;\;\;\;for\;0 < x < 1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Let A be the event that the tack will land with its point up on all three tosses.

Then, \({\rm P}\left( {A|X = x} \right) = {x^3}\)

Hence,

\({\rm P}\left( A \right) = \int\limits_0^1 {{x^3} \times 2\left( {1 - x} \right)dx} \)

\(\begin{aligned}{}{\rm P}\left( A \right) &= 2\int\limits_0^1 {\left( {1 - x} \right){x^3}dx} \\ &= 2\int\limits_0^1 \begin{aligned}{l}\left( {{x^3} - {x^4}} \right)dx\\\end{aligned} \\ &= \frac{1}{{10}}\end{aligned}\)

Hence Probability that the tack will land with its point up on all three tosses is \(\frac{1}{{10}}\)