Chapter 13.10, Problem 24PPS

To determine

The probability of the same kind bar Luis and Granola receive.

Answer to Problem 24PPS

The probability of the same kind bar Luis and Granola receive is P (Pair ofgranola bars) = $\frac{49}{153}$ .

Explanation of Solution

Given:

Cinnamon Apple granola bars: $6$

PeanutApple granola bars: $4$

Oatmeal raisin granola bars: $8$

Calculation:

Fraction Form:

It is given the box has $6$ apple cinnamon granola bars, $4$ peanut granola bars and $8$ oatmeal raisin granola bars.

So no matter if cinnamon, peanut or oatmeal raisin to get a pair, so pick 2 simultaneously. The probability of getting $2$ granola bars consecutively form is given as,

P (Pair of Apple Cinnamon):

For this take the number of total of apple cinnamon and total number of granola bars.

P (First bar is Apple Cinnamon) = $\frac{6}{18}$

Now remove the $1$ bar apple cinnamon from the total number of apple cinnamon.

P (Second bar is Apple Cinnamon) = $\frac{5}{17}$

P (Pair of Apple Cinnamon) = P (First) * P (Second)

  $\begin{array}{l}=\frac{6}{18}\times \frac{5}{17}\\ =\left(\frac{6.5}{18.17}\right)\\ =\frac{30}{306}\end{array}$

So the P (Pair of Apple Cinnamon) $=\frac{30}{306}$

P (Pair of peanut bars)

For this take the number of total of peanut bars and total number of granola bars.

P (First bar ispeanut bars) = $\frac{4}{18}$

Now remove the $1$ peanut barsfrom total number of peanut bars, the probability is,

P (Second bar ispeanut bars) = $\frac{3}{17}$

P (Pair of peanut bars) = P (First) * P (Second)

  $\begin{array}{l}=\frac{4}{18}\times \frac{3}{17}\\ =\left(\frac{4.3}{18.17}\right)\\ =\frac{12}{306}\end{array}$

So the P (Pair of peanut bars) $=\frac{12}{306}$

P (Pair of Oatmeal Raisin):

For this take the number of total of Oatmeal Raisinand total number of granola bars.

P (First bar is Oatmeal Raisin) = $\frac{8}{18}$

Now remove the $1$ Oatmeal Raisin from the total of Oatmeal Raisin, the probability,

P (Second bar is Oatmeal Raisin) = $\frac{7}{17}$

P (Pair of Oatmeal Raisin) = P (First) * P (Second)

  $\begin{array}{l}=\frac{10}{18}\times \frac{9}{17}\\ =\left(\frac{8.7}{18.17}\right)\\ =\frac{56}{306}\end{array}$

So the P (Pair of Oatmeal Raisin) $=\frac{56}{306}$

Now, as there are $3$ events cannot happed at the exact same time, so this is the example of the mutually exclusive event. So the probability of a picking any pair of socks is s follows,

P (Pair of granola bars):

P (Pair ofgranola bars) = P (Apple cinnamon) + P (Peanut) + P (Oatmeal Raisin)

  $\begin{array}{l}=\frac{30}{306}+\frac{12}{306}+\frac{56}{306}\\ =\left(\frac{30+12+56}{306}\right)\\ =\left(\frac{98}{306}\right)\\ =\frac{49}{153}\end{array}$

Conclusion:

Hence, P (Pair ofgranola bars)= $\frac{49}{153}$