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, localid="1649537434462" $\mathrm{P}\left(\mathrm{H}\right)=\frac{2}{3}$andlocalid="1649537225213" $\mathrm{P}\left(\mathrm{T}\right)=\frac{1}{3}$wherelocalid="1649537596463" $\mathrm{H}$is heads andlocalid="1649537603746" $\mathrm{T}$is tails .

Find localid="1649537612097" $\mathrm{P}$( tossing a Head on the coin AND a Red bead )

a. $\frac{2}{3}$

b.$\frac{5}{15}$

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

d. $\frac{5}{36}$

Short answer

$\mathrm{P}(\mathrm{tossing}\mathrm{a}\mathrm{Head}\mathrm{on}\mathrm{the}\mathrm{coin}\mathrm{AND}\mathrm{a}\mathrm{red}\mathrm{bead})$is$\frac{6}{36}$,So answer c is correct.

Step-by-step solution

Given data

From tree diagram,

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

Probability of tossing head on the coin with red bead$\mathrm{P}\left(\mathrm{R}\right|\mathrm{H})$$=$$\frac{3}{12}$.

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

Calculation of P(tossing a Head on the coin AND a red bean)

Probability is,

$\mathrm{P}(\mathrm{tossing}\mathrm{a}\mathrm{Head}\mathrm{on}\mathrm{the}\mathrm{coin}\mathrm{AND}\mathrm{a}\mathrm{red}\mathrm{bean})$$=>$$\mathrm{P}(\mathrm{H}\mathrm{AND}\hspace{0.17em}\mathrm{R})$

$=\mathrm{P}(\mathrm{Red}\mathrm{bead}|\mathrm{tossing}\mathrm{a}\mathrm{Head}\mathrm{on}\mathrm{the}\mathrm{coin})\mathrm{P}(\mathrm{tossing}\mathrm{a}\mathrm{Head}\mathrm{on}\mathrm{the}\mathrm{coin})$

$=\mathrm{P}\left(\mathrm{R}\right|\mathrm{H})\times \mathrm{P}(\mathrm{H})$

$=\frac{3}{12}\times \frac{2}{3}$

$=\frac{6}{36}$