Chapter 13.10, Problem 19PPS

To determine

The probability ofeach outcome of the marbles.

Three green marbles in a row.

Answer to Problem 19PPS

The probability ofthree green marbles in a row is= $\frac{1}{91}$

Explanation of Solution

Given:

Blue: $5$

Red: $3$

Green: $4$

Yellow: $2$

Total Marbles: $5+3+4+2=14$

Calculation:

Fraction Form:

So calculate the probability of the $3$ green marbles in a row in this form is as calculate the as the follows, so take the number of green marbles and the total number of the marbles.

P (First marble is green) = $\frac{4}{14}=\frac{2}{7}$

Now remove the $1$ marble from green marbles, the probability is,

P (Second marble is green) = $\frac{3}{13}$

Now take the number green marbles after removed the $1$ more green marbles,

P (Third candy is blue) = $\frac{2}{12}=\frac{1}{6}$

P (Third Green Marbles) = P (First) * P (Second)*P(Third)

  $\begin{array}{l}=\frac{2}{7}\times \frac{3}{13}\times \frac{1}{6}\\ =\left(\frac{\mathrm{2.3.1}}{\mathrm{7.13.6}}\right)\\ =\frac{6}{546}\\ =\frac{1}{91}\end{array}$

Conclusion:

Hence, P (three green marbles)= $\frac{1}{91}$