Represent a multi-stage finite sample space

What must stay distinct when a sample space records successive stages?

Codes 1,2,3,4 mean RR,RB,BR,BB. The stages are independent and the second-stage probability of R is one third. Vary the first-stage probability p. Table probabilities are rounded to 2 decimal places and event readouts to 3, with trailing zeroes omitted. Rounded rows may not add to 1; the total uses exact probabilities. Compare the reversed paths RB and BR; they are distinct outcomes and need not have the same probability

When the first-stage probability of R is one half, are all four ordered paths equally likely?

Product sample spaces

A multi-stage sample space contains one ordered tuple for every choice at every stage. Count all tuples first, then count the tuples that satisfy the event

1. 1Keep stage order The position of a result in the tuple records the stage where it occurred \(\displaystyle (A,B)\ne(B,A)\)
2. 2Multiply stage choices Multiply the number of choices at each successive stage to find the total \(\displaystyle N=3\times4=12\)
3. 3Count favourable tuples Count the ordered tuples that satisfy the event condition \(\displaystyle F=3\)
4. 4Divide by total outcomes For equally likely tuples, divide the favourable count by the full sample-space count \(\displaystyle \Pr(E)=\frac{F}{N}=\frac{3}{12}=\frac{1}{4}\)

Complete product space

\(\displaystyle n_1=3,\quad n_2=4,\quad\text{every pair is possible}\)

A two-stage experiment has 3 first-stage outcomes and 4 second-stage outcomes. Find the total number of ordered outcomes

Complete product space

\(\displaystyle |S|=12,\quad |E|=3,\quad\text{all 12 ordered outcomes equally likely}\)

The event E consists of 3 particular ordered outcomes among these 12 outcomes. Find the exact probability of E.

Adding stage counts. One of these lines is right. Which one?

Read two-stage branch probabilities

Use two independent stages, each labelled R or B. The table gives the probability of each label at each stage. From either first-stage branch, use the same second-stage probabilities. Record both stage labels before multiplying

Independent stages and event

\(\displaystyle n_1=2,\quad n_2=2,\quad S=\{RR,RB,BR,BB\}\)

How many ordered paths are in this two-stage product space?

Independent stages and event

\(\displaystyle \Pr(R_1)=\frac23,\quad\Pr(R_2)=\frac14,\quad\text{stages independent}\)

Find the probability of the ordered path RR

Independent stages and event

\(\displaystyle \Pr(R_1)=\frac23,\quad\Pr(B_2)=\frac34,\quad\text{stages independent}\)

Find the probability of the ordered path RB

Independent stages and event

\(\displaystyle \Pr(B_1)=\frac13,\quad\Pr(R_2)=\frac14,\quad\text{stages independent}\)

Find the probability of the reversed ordered path BR

Check every ordered path and its mass

The four masses sum to one. RB and BR both contain one R but their probabilities differ. They can be grouped for an event without being merged into one elementary outcome

Exactly one R occurs on two of the four paths. Can its probability be found by counting favourable paths here?

Independent stages and event

\(\displaystyle \Pr(RB)=\frac12,\quad\Pr(BR)=\frac1{12}\)

Find the probability of exactly one R.

Independent stages and event

\(\displaystyle \Pr(BB)=\frac14\)

Find the probability of at least one R.

Distinguish exactly one from matching results

Matching results occur on RR or BB, with total mass 1/6+1/4=5/12. Exactly one R and matching results partition this two-stage space

Extend every path to a third independent stage

Retain the first two stages and add a third independent stage with Pr(R)=3/5 and Pr(B)=2/5. Every two-stage path has two extensions. A condition on the first two entries can leave the third entry unrestricted

Independent stages and event

\(\displaystyle \Pr(R_1B_2)=\frac12,\quad\Pr(R_3)+\Pr(B_3)=\frac35+\frac25=1\)

Find the probability that the first two results are R then B, with either third result

Independent stages and event

\(\displaystyle \Pr(R_1)=\frac23,\quad\Pr(B_2)=\frac34,\quad\Pr(R_3)=\frac35,\quad\text{independent}\)

Find the probability of the single ordered path RBR

Which event describes first two results R then B, rather than the one complete path RBR?

Independent stages and event

\(\displaystyle \Pr(RBR)=\frac3{10},\quad\Pr(BRB)=\frac1{30}\)

Find the probability that each result differs from the preceding one

Independent stages and event

\(\displaystyle \Pr(R_1)=\frac23,\quad\Pr(B_1)=\frac13,\quad\Pr(R_3)=\frac35,\quad\Pr(B_3)=\frac25\)

Find the probability that the first and last results differ, with either middle result

Do not replace one event condition by another

With three binary stages, adjacent-different paths RBR and BRB have matching first and last entries. The first/last-different event instead contains RRB,RBB,BRR,BBR. Translate the words into a set before calculating

Construct the full tree and count distribution, ungraded

Each of three independent stages has outcomes R and B. Draw the complete tree with branch probabilities. List all eight ordered paths and their masses. Let X count the R results; group every path by X and write the complete distribution

Independent stages and event

\(\displaystyle \Pr(B_1)=\frac13,\quad\Pr(B_2)=\frac34,\quad\Pr(B_3)=\frac25,\quad\text{independent}\)

Let X count the R results in three stages with outcomes R or B. After constructing the complete table, enter Pr(X=0)

Independent stages and event

\(\displaystyle \Pr(RBB)=\frac15,\quad\Pr(BRB)=\frac1{30},\quad\Pr(BBR)=\frac3{20}\)

Let X count the R results in three stages with outcomes R or B. Enter Pr(X=1)

Independent stages and event

\(\displaystyle \Pr(RRB)=\frac1{15},\quad\Pr(RBR)=\frac3{10},\quad\Pr(BRR)=\frac1{20}\)

Let X count the R results in three stages with outcomes R or B. Enter Pr(X=2)

Independent stages and event

\(\displaystyle \Pr(R_1)=\frac23,\quad\Pr(R_2)=\frac14,\quad\Pr(R_3)=\frac35,\quad\text{independent}\)

Let X count the R results in three stages with outcomes R or B. Enter Pr(X=3)

Compare every path before checking the grouped total

The path list below gives all eight terminal masses. Grouping by the number of R results gives probabilities of one tenth, twenty-three sixtieths, five twelfths and one tenth for zero, one, two and three R results, respectively. Their sum is one. The stage probabilities differ, so this is direct enumeration, not an identical-trial binomial formula

Check the representation, event and probability separately

A complete list, table or tree retains every ordered outcome exactly once. Match the event wording to the correct subset. For independent stages multiply branch probabilities along each path, then add disjoint favourable path masses. Only an equiprobable space permits the count ratio