Probability & Statistics 1

Mathematics9709

Probability & Statistics 1

Mathematics9709

Building a distribution from outcomes

A probability distribution gives every possible value of $X$ and the probability attached to it.

One distribution, two views

x012

P(X=x)1/41/21/4

1/40

1/21

1/42

1/4 + 1/2 + 1/4 = 1 ✓

Every possible value appears once, directly above its probability. The probabilities are non-negative and add to 1.

A complete table passes three checks

Every possible value appears once.
Every probability is between 0 and 1 inclusive.
All probabilities add to 1.

\sum P(X=x)=1

A vertical line graph is another view of the same distribution. Each line stands at one possible value and its height is the probability. The gaps matter: this is not a continuous histogram.

Map outcomes, then pool equal values

To pool outcomes means to add the probabilities of different outcomes that produce the same value of $X$ .

Worked example

Two fair coin tosses

The ordered outcomes HH, HT, TH and TT are equally likely.

$X=0$ comes from TT, so its probability is $1/4$ .
$X=1$ comes from HT or TH, so its probability is $2/4=1/2$ .
$X=2$ comes from HH, so its probability is $1/4$ .

x	P(X = x)
0	1/4
1	1/2
2	1/4

x	0	1	2
P(X = x)	1/4	1/2	1/4

Common mistake

HT and TH are different ordered outcomes. Do not lose one of them, and do not give

X=1

two separate columns.

When outcomes are not equally likely, use path probabilities

Suppose three independent coins have head probabilities $1/3$ , $1/4$ and $1/5$ . Independent means one result does not change the probabilities for the others. For one path, multiply. For several paths giving the same head count, add. (9709/51/M/J/24 Q6(a))

Worked example

Exactly one head

P(HTT)=\frac13\frac34\frac45=\frac{12}{60}

P(THT)=\frac23\frac14\frac45=\frac8{60}

P(TTH)=\frac23\frac34\frac15=\frac6{60}

The three path numerators add to 26.

P(X=1)=\frac{26}{60}=\frac{13}{30}

x	P(X = x)
0	2/5
1	13/30
2	3/20
3	1/60

x	0	1	2	3
P(X = x)	2/5	13/30	3/20	1/60

Use an equally likely outcome count only when the elementary outcomes really are equally likely. Otherwise calculate each path.

Your turnIndependent check [4 marks]

A fair spinner labelled 1, 2 and 3 is spun twice. Let $S$ be the sum. Draw up the distribution of $S$ .

Show worked answer

The nine ordered outcomes are equally likely. Their sums have frequencies 1, 2, 3, 2 and 1.

Answer

x	P(X = x)
2	1/9
3	2/9
4	3/9
5	2/9
6	1/9

x	2	3	4	5	6
P(X = x)	1/9	2/9	3/9	2/9	1/9

PreviousDiscrete random variables NextDistributions from selections and stopping rules

Building a distribution from outcomes

A probability distribution gives every possible value of $X$ and the probability attached to it.

One distribution, two views

x012

P(X=x)1/41/21/4

1/40

1/21

1/42

1/4 + 1/2 + 1/4 = 1 ✓

Every possible value appears once, directly above its probability. The probabilities are non-negative and add to 1.

A complete table passes three checks

Every possible value appears once.
Every probability is between 0 and 1 inclusive.
All probabilities add to 1.

\sum P(X=x)=1

A vertical line graph is another view of the same distribution. Each line stands at one possible value and its height is the probability. The gaps matter: this is not a continuous histogram.

Map outcomes, then pool equal values

To pool outcomes means to add the probabilities of different outcomes that produce the same value of $X$ .

Worked example

Two fair coin tosses

The ordered outcomes HH, HT, TH and TT are equally likely.

$X=0$ comes from TT, so its probability is $1/4$ .
$X=1$ comes from HT or TH, so its probability is $2/4=1/2$ .
$X=2$ comes from HH, so its probability is $1/4$ .

x	P(X = x)
0	1/4
1	1/2
2	1/4

x	0	1	2
P(X = x)	1/4	1/2	1/4

Common mistake

HT and TH are different ordered outcomes. Do not lose one of them, and do not give

X=1

two separate columns.

When outcomes are not equally likely, use path probabilities

Worked example

Exactly one head

P(HTT)=\frac13\frac34\frac45=\frac{12}{60}

P(THT)=\frac23\frac14\frac45=\frac8{60}

P(TTH)=\frac23\frac34\frac15=\frac6{60}

The three path numerators add to 26.

P(X=1)=\frac{26}{60}=\frac{13}{30}

x	P(X = x)
0	2/5
1	13/30
2	3/20
3	1/60

x	0	1	2	3
P(X = x)	2/5	13/30	3/20	1/60

Use an equally likely outcome count only when the elementary outcomes really are equally likely. Otherwise calculate each path.

Your turnIndependent check [4 marks]

A fair spinner labelled 1, 2 and 3 is spun twice. Let $S$ be the sum. Draw up the distribution of $S$ .

Show worked answer

The nine ordered outcomes are equally likely. Their sums have frequencies 1, 2, 3, 2 and 1.

Answer

x	P(X = x)
2	1/9
3	2/9
4	3/9
5	2/9
6	1/9

x	2	3	4	5	6
P(X = x)	1/9	2/9	3/9	2/9	1/9

PreviousDiscrete random variables NextDistributions from selections and stopping rules