Question

A family has a day of 7 activities planned: shopping, picnic, hiking, swimming, bike ride, video games, and movie. To make it more adventurous they decide to randomly pick the order of the activities out of a hat. Find the probability that bike ride and movie are chosen consecutively, in either order.

Answer 1

`Pr= (1)/(21)`

Explanation:

Given

`n(S) = 7` --- number of games

Required

Probability of bike and movie in consecutive order

This probability is represented as:

`Pr = P(Bike\ and\ Movie) \ or\ P(Movie\ or\ Bike)`

So, we have:

`Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)`

The probability is an illustration of selection without replacement;

So, we have:

`P(Bike\ and\ Movie) = P(Bike) * P(Movie)`

`P(Bike\ and\ Movie) = (n(Bike))/(n(S)) * (n(Movie))/(n(S) - 1)` ---- without replacement

Bike and Movie appear in the game list 1 time.

So, the equation becomes

`P(Bike\ and\ Movie) = (1)/(7) * (1)/(7 - 1)`

`P(Bike\ and\ Movie) = (1)/(7) * (1)/(6)`

`P(Bike\ and\ Movie) = (1)/(42)`

Similarly,

`P(Movie\ and\ Bike) = (1)/(42)`

So, we have:

`Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)`

`Pr= (1)/(42)+(1)/(42)`

Take LCM

`Pr= (1+1)/(42)`

`Pr= (2)/(42)`

`Pr= (1)/(21)`