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

Answer:

$Pr= \frac{1}{21}$

Step-by-step 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:

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

$P(Bike\ and\ Movie) = \frac{n(Bike)}{n(S)} * \frac{n(Movie)}{n(S) - 1}$ ---- without replacement

Bike and Movie appear in the game list 1 time.

So, the equation becomes

$P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{7 - 1}$

$P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{6}$

$P(Bike\ and\ Movie) = \frac{1}{42}$

Similarly,

$P(Movie\ and\ Bike) = \frac{1}{42}$

So, we have:

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

$Pr= \frac{1}{42}+\frac{1}{42}$

Take LCM

$Pr= \frac{1+1}{42}$

$Pr= \frac{2}{42}$

$Pr= \frac{1}{21}$