Probability & Statistics 1

Mathematics9709

Probability & Statistics 1

Mathematics9709

Distributions from selections and stopping rules

Selection questions split into two forms: a fixed-size group, or an ordered process that continues until something happens.

Five coins give 10 equally likely pairs

$5$2a$2b$1a$1b

x23467

P(X=x)1/104/101/102/102/10

pair counts: 1 + 4 + 1 + 2 + 2 = 10

Treat the five physical coins as distinct when counting pairs, even when two coins have the same value. Then pool pairs that give the same total.

Fixed-size selections: count groups

Jasmine selects two coins from one $5 coin, two $2 coins and two $1 coins. The physical coins are distinct, even when their printed values match. There are ${}^5C_2=10$ equally likely unordered pairs. (9709/52/M/J/24 Q5)

List the possible totals: 2, 3, 4, 6 and 7.
Count the physical pairs giving each total.
Divide every count by the same total, 10.

x	P(X = x)
2	1/10
3	4/10
4	1/10
6	2/10
7	2/10

x	2	3	4	6	7
P(X = x)	1/10	4/10	1/10	2/10	2/10

Common mistake

The two $2 coins are not one selectable object. Label them temporarily while counting, then pool pairs that have the same total value.

Count each table cell with combinations

If 2 counters are selected from 3 red and 5 blue counters and $R$ is the number of red counters, then

{}^8C_2=28

P(R=r)=\frac{{}^3C_r\,{}^5C_{2-r}}{28}

Use $r=0,1,2$ to produce the three table cells.

The denominator counts all two-counter groups. The numerator chooses exactly $r$ red counters and the remaining $2-r$ blue counters. Every table cell must use the same sample space.

Examiner note

State the possible values before calculating. A forced minimum or maximum can remove values that look possible at first.

Stopping processes: multiply in order

Without replacement means a selected object is not returned, so the numbers and probabilities change after every draw. If $X$ is the draw on which the first green counter appears, then $X=k$ means: non-green on the first $k-1$ draws, then green.

Worked example

Three non-green and two green counters

Draw without replacement until the first green. The latest possible stop is draw 4, because only three non-green counters exist.

P(X=1)=\frac25

P(X=2)=\frac35\frac24=\frac3{10}

P(X=3)=\frac35\frac24\frac23=\frac15

P(X=4)=\frac35\frac24\frac13=\frac1{10}

Answer

\frac25+\frac3{10}+\frac15+\frac1{10}=1

Common mistake

Do not use one fixed probability on every draw. After a non-green counter is removed, both the numerator and denominator may change.

Your turnIndependent check [4 marks]

Two counters are selected from 3 red and 2 blue counters. Let $B$ be the number of blue counters. Draw up the distribution of $B$ .

Show worked answer

There are ${}^5C_2=10$ equally likely groups. Count the groups with 0, 1 and 2 blue counters.

Answer

x	P(X = x)
0	3/10
1	6/10
2	1/10

x	0	1	2
P(X = x)	3/10	6/10	1/10

PreviousBuilding a distribution from outcomes NextCompleting and using a distribution

Distributions from selections and stopping rules

Selection questions split into two forms: a fixed-size group, or an ordered process that continues until something happens.

Five coins give 10 equally likely pairs

$5$2a$2b$1a$1b

x23467

P(X=x)1/104/101/102/102/10

pair counts: 1 + 4 + 1 + 2 + 2 = 10

Treat the five physical coins as distinct when counting pairs, even when two coins have the same value. Then pool pairs that give the same total.

Fixed-size selections: count groups

List the possible totals: 2, 3, 4, 6 and 7.
Count the physical pairs giving each total.
Divide every count by the same total, 10.

x	P(X = x)
2	1/10
3	4/10
4	1/10
6	2/10
7	2/10

x	2	3	4	6	7
P(X = x)	1/10	4/10	1/10	2/10	2/10

Common mistake

The two $2 coins are not one selectable object. Label them temporarily while counting, then pool pairs that have the same total value.

Count each table cell with combinations

If 2 counters are selected from 3 red and 5 blue counters and $R$ is the number of red counters, then

{}^8C_2=28

P(R=r)=\frac{{}^3C_r\,{}^5C_{2-r}}{28}

Use $r=0,1,2$ to produce the three table cells.

The denominator counts all two-counter groups. The numerator chooses exactly $r$ red counters and the remaining $2-r$ blue counters. Every table cell must use the same sample space.

Examiner note

State the possible values before calculating. A forced minimum or maximum can remove values that look possible at first.

Stopping processes: multiply in order

Worked example

Three non-green and two green counters

Draw without replacement until the first green. The latest possible stop is draw 4, because only three non-green counters exist.

P(X=1)=\frac25

P(X=2)=\frac35\frac24=\frac3{10}

P(X=3)=\frac35\frac24\frac23=\frac15

P(X=4)=\frac35\frac24\frac13=\frac1{10}

Answer

\frac25+\frac3{10}+\frac15+\frac1{10}=1

Common mistake

Do not use one fixed probability on every draw. After a non-green counter is removed, both the numerator and denominator may change.

Your turnIndependent check [4 marks]

Two counters are selected from 3 red and 2 blue counters. Let $B$ be the number of blue counters. Draw up the distribution of $B$ .

Show worked answer

There are ${}^5C_2=10$ equally likely groups. Count the groups with 0, 1 and 2 blue counters.

Answer

x	P(X = x)
0	3/10
1	6/10
2	1/10

x	0	1	2
P(X = x)	3/10	6/10	1/10

PreviousBuilding a distribution from outcomes NextCompleting and using a distribution