Conditional Probability

Worked example

1. 1

   $\text{type A}=38+47=85$ — “One of the type A phones” means only type A can be chosen. Add along the row: 85 phones is the new total.

2. 2

   $P(128\text{ GB})=\frac{47}{85}$ — 47 of those 85 type A phones are 128 GB. The group you are given goes on the bottom.

3. 3

   $\text{128 GB}=47+41=88$ — Now the question says “given 128 GB”. Add down the column: 88 phones are 128 GB.

4. 4

   $P(\text{type B})=\frac{41}{88}$ — 41 of those 88 are type B, so 88 is the bottom this time.

Worked example

1. 1

   $t>24:13+10=23$ — More than 24 minutes means the last two rows. Add them: 23 shoppers is the group you are given.

2. 2

   $t>36:10$ — Inside that group, only the last row took more than 36 minutes.

3. 3

   $P=\frac{10}{23}$ — The part you want over the group you are given. 10 and 23 share no factor, so it does not simplify.

Worked example

1. 1

   $n(S)=6+4+2+1=13$ — Add every region inside the S circle, including the overlaps. These 13 swimmers are the only ones who can be picked.

2. 2

   $n(S\cap T)=2+1=3$ — Inside S, find the regions that are also in T: “T and S only” and “all three”. That is 3.

3. 3

   $P(T\text{ given }S)=\frac{3}{13}$ — The part you want over the circle you are given.

Worked example

1. 1

   $P(\text{rain})=0.3,P(\text{rain}\cap \text{wind})=0.12$ — Read the two probabilities. The day is given to be rainy, so rain is B.

2. 2

   $P(\text{wind}\mid \text{rain})=\frac{0.12}{0.3}$ — Put “both” on top and the given event on the bottom.

3. 3

   $=0.4$ — 0.12 ÷ 0.3 = 1.2 ÷ 3 = 0.4. It is bigger than 0.12 because only rainy days count now.

Worked example

1. 1

   $3\times 3=9\text{cells}$ — Every first number goes with every second number. Each of the 9 cells is equally likely.

2. 2

   $3,5,5,7,7$ — “Given the total is odd”: keep only the odd cells. There are 5, so 5 goes on the bottom.

3. 3

   $1+2,2+5,3+2$ — Inside those 5 cells, find the ones where one of the numbers picked is a 2. There are 3.

4. 4

   $P=\frac{3}{5}$ — The cells you want over the cells you are given.

Worked example

1. 1

   $\text{group}=12$ — Only the 12 chess players can be picked, so 12 is the first bottom. The other 18 members do not count.

2. 2

   $P(\text{1st draughts})=\frac{8}{12}$ — 8 of the 12 chess players also play draughts.

3. 3

   $P(\text{2nd draughts})=\frac{7}{11}$ — The same person cannot be picked twice. One draughts player is gone, so 7 are left out of 11.

4. 4

   $\frac{8}{12}\times \frac{7}{11}=\frac{56}{132}$ — Both must happen, so multiply along the path.

5. 5

   $=\frac{14}{33}$ — Divide top and bottom by 4 to simplify.

Worked example

1. 1

   $P(\text{W, W})=0.6\times 0.7=0.42$ — Win the first race (0.6), then win again. After a win the branch is 0.7. Multiply along the path.

2. 2

   $P(\text{W, L})=0.6\times 0.3=0.18$ — Win, then lose. After a win, losing is 1 − 0.7 = 0.3, because the two branches add to 1.

3. 3

   $P(\text{L, W})=0.4\times 0.4=0.16$ — Lose the first (1 − 0.6 = 0.4), then win. After a loss the branch is 0.4.

4. 4

   $\text{exactly one}=0.18+0.16=0.34$ — Win then lose, and lose then win, both have exactly one win. Either path will do, so add them.