Question

Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red$\left(\mathrm{R}\right)$,four yellow$\left(\mathrm{Y}\right)$and five blue$\left(\mathrm{B}\right)$beads. For the coin, and$\mathrm{P}\left(\mathrm{H}\right)=\frac{2}{3}$and$\mathrm{P}\left(\mathrm{T}\right)=\frac{1}{3}$where$\mathrm{H}$is heads and$\mathrm{T}$is tails .

Find $\mathrm{P}$(Blue bead).

a.$\frac{15}{36}$

b.$\frac{10}{36}$

c.$\frac{10}{12}$

d.$\frac{6}{36}$

Short answer

Possibility of five blue bead is$\frac{10}{12}$,So answer c is correct.

Step-by-step solution

Given data

From tree diagram,

Probability of tossing Head on the coin is $\mathrm{P}\left(\mathrm{H}\right)=\frac{2}{3}$.

Probability of tossing Tail on the coin is $\mathrm{P}\left(\mathrm{T}\right)=\frac{1}{3}$.

Probability of tossing head on the coin with five blue beads is $\mathrm{P}\left(\mathrm{B}\right|\mathrm{H})=\frac{5}{12}$.

Probability of tossing tail on the coin with five blue beads is$\mathrm{P}\left(\mathrm{B}\right|\mathrm{T})=\frac{5}{12}$

Calculation for possibility of five blue beads

Possibility is,

$\mathrm{P}(\mathrm{Blue}\mathrm{bead})$

$=\mathrm{P}(\mathrm{Blue}\mathrm{bead}|\mathrm{tossing}\mathrm{a}\mathrm{Head}\mathrm{on}\mathrm{coil})+\mathrm{P}(\mathrm{Blue}\mathrm{bead}|\mathrm{tossing}\mathrm{a}\mathrm{tail}\mathrm{on}\mathrm{coil})$

$=\mathrm{P}\left(\mathrm{B}\right|\mathrm{H})+\mathrm{P}(\mathrm{B}\left|\mathrm{T}\right)$

$=\frac{5}{12}+\frac{5}{12}$

$=\frac{10}{12}$