4.6 Probability topics: tree diagrams (optional)  (Page 8/3)

• $\mathrm{B1R1}$
• $\mathrm{B1R2}$
• $\mathrm{B1R3}$
• $\mathrm{B2R1}$
• $\mathrm{B2R2}$
• $\mathrm{B2R3}$
• $\mathrm{B3R1}$
• $\mathrm{B3R2}$
• $\mathrm{B3R3}$
• $\mathrm{B4R1}$
• $\mathrm{B4R2}$
• $\mathrm{B4R3}$
• $\mathrm{B5R1}$
• $\mathrm{B5R2}$
• $\mathrm{B5R3}$
• $\mathrm{B6R1}$
• $\mathrm{B6R2}$
• $\mathrm{B6R3}$
• $\mathrm{B7R1}$
• $\mathrm{B7R2}$
• $\mathrm{B7R3}$
• $\mathrm{B8R1}$
• $\mathrm{B8R2}$
• $\mathrm{B8R3}$

$\text{P(RR)}=\frac{3}{11}\cdot \frac{3}{11}=\frac{9}{121}$

$\text{P(RB OR BR)}=\frac{3}{11}\cdot \frac{8}{11}+\frac{8}{11}\cdot \frac{3}{11}=\frac{48}{121}$

$\text{P(R on 1st draw AND B on 2nd draw)}=\text{P(RB)}=\frac{3}{11}\cdot \frac{8}{11}=\frac{24}{121}$

$\text{P(R on 2nd draw given B on 1st draw)}=\text{P(R on 2nd | B on 1st)}=\frac{24}{88}=\frac{3}{11}$

This problem is a conditional. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are $24 + 64 = 88$ possible outcomes (24 $\mathrm{BR}$ and 64 $\mathrm{BB}$ ). Twenty-four of the 88 possible outcomes are $\mathrm{BR}$ . $\frac{24}{88}=\frac{3}{11}$ .

$\text{P(BB)} = \frac{64}{121}$

$\text{P(B on 2nd draw | R on 1st draw)} = \frac{8}{11}$

There are $9 + 24$ outcomes that have $R$ on the first draw (9 $\mathrm{RR}$ and 24 $\mathrm{RB}$ ). The sample space is then $9 + 24 = 33$ . Twenty-four of the 33 outcomes have $B$ on the second draw. The probability is then $\frac{24}{33}$ .

$\text{P(RR)}=\frac{3}{11}\cdot \frac{2}{10}=\frac{6}{110}$

$\text{P(RB or BR)}=\frac{3}{11}\cdot \frac{8}{10}+$ $\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)$ $=\frac{48}{110}$

$\text{P(R on 2d | B on 1st)} = \frac{3}{10}$

$\text{P(R on 1st and B on 2nd)} = \text{P(RB)} =$ $\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)$ $= \frac{24}{110}$

$\text{P(BB)} = \frac{8}{11} \cdot \frac{7}{10}$

There are $6 + 24$ outcomes that have $R$ on the first draw (6 $\mathrm{RR}$ and 24 $\mathrm{RB}$ ). The 6 and the 24 are frequencies. They are also the numerators of the fractions $\frac{6}{110}$ and $\frac{24}{110}$ . The sample space is no longer 110 but $6 + 24 = 30$ . Twenty-four of the 30 outcomes have $B$ on the second draw. The probability is then $\frac{24}{30}$ . Did you get this answer?