Tree Diagrams Flashcards

All 8 cards in this deck

True or false? On a tree diagram for two counters taken from a bag without replacement, the second pair of branches shows the same probabilities as the first pair.

False. One counter has gone, so the denominator is one less and the probabilities change — using the same denominator is a common error.

What are the steps to complete a probability tree diagram for two selections without replacement?

e.g. a bag holds 2 red and 3 blue counters; two counters are taken

1. First pair of branches from the starting amounts: $\frac{2}{5}$ red, $\frac{3}{5}$ blue.
2. Second branches: total drops to 4 and the colour taken drops by 1, e.g. after red: $\frac{1}{4}$ red, $\frac{3}{4}$ blue.
3. Check every pair of branches totals 1, e.g. after blue: $\frac{2}{4}+\frac{2}{4}=1$.

How do you find the probability of a combination of outcomes from a probability tree diagram?

e.g. P(spinner A lands red) $=0.25$, P(spinner B lands red) $=0.6$; find P(both land on red)

Multiply the probabilities along the branches: $0.25 \times 0.6 = 0.15$.

How do you find the probability of 'at least one' from a tree diagram?

e.g. P(train late) $=0.15$ on each of two days; find P(late on at least one day)

Use $1 - P(\text{neither})$, or add the probabilities of all the end outcomes containing it: $1 - 0.85\times0.85 = 0.2775$.

How do you find the probability on a second branch of a tree diagram when you know the first-branch probability and the probability of the combined outcome?

e.g. P(rain on Monday) $=0.2$ and P(rain on both Monday and Tuesday) $=0.06$

Divide the combined probability by the first-branch probability: $0.06 \div 0.2 = 0.3$ on the 'rain Tuesday' branch that follows 'rain Monday'.

What are the steps to find an unknown branch probability $x$ from a given overall probability on a tree diagram?

e.g. P(win) $=x$ on each of two games, and P(win both games) $=0.36$

1. Multiply along the branches in terms of $x$: $x \times x = x^2$
2. Set this equal to the given probability: $x^2 = 0.36$
3. Solve, rejecting any value outside $0 \le x \le 1$: $x = 0.6$

What assumption is being made when a tree diagram shows the same probabilities on the second pair of branches as on the first pair?

e.g. P(train late) $=0.15$ on the branches for both day 1 and day 2

That the two events are independent: whether the train is late on day 1 does not affect the probability that it is late on day 2.

True or false? You may work out P(A and B) as P(A) $\times$ P(B) only if A and B are independent.

True. If they are not independent, the second branch must show the conditional probability of B given A, as on a 'without replacement' tree.