# 3.5 Tree and venn diagrams  (Page 2/10)

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## Try it

In a standard deck, there are 52 cards. Twelve cards are face cards ( F ) and 40 cards are not face cards ( N ). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

1. Find P ( FN OR NF ).
2. Find P ( N | F ).
3. Find P (at most one face card).
Hint: "At most one face card" means zero or one face card.
4. Find P (at least on face card).
Hint: "At least one face card" means one or two face cards.
1. P ( FN OR NF ) =
2. P ( N | F ) = $\frac{40}{51}$
3. P (at most one face card) = = $\frac{2,520}{2,652}$
4. P (at least one face card) = = $\frac{\text{1,092}}{\text{2,652}}$

A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

1. What is the probability that both kittens are tabby?
a. $\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$ b. $\left(\frac{4}{9}\right)\left(\frac{4}{9}\right)$ c. $\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)$ d. $\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)$
2. What is the probability that one kitten of each coloring is selected?
a. $\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)$ b. $\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)$ c. $\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right)$ d. $\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)$
3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
4. What is the probability of choosing two kittens of the same color?

a. c, b. d, c. $\frac{4}{8}$ , d. $\frac{32}{72}$

## Try it

Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?

$\left(\frac{4}{7}\right)\left(\frac{3}{6}\right)$ + $\left(\frac{3}{7}\right)\left(\frac{4}{6}\right)$

## Venn diagram

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.

Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A  OR  B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:

## Try it

Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C AND P = {blue} and C OR P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.

Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = { TT , TH } and B = { TT , HT }. Therefore, A AND B = { TT }. A OR B = { TH , TT , HT }.

The sample space when you flip two fair coins is X = { HH , HT , TH , TT }. The outcome HH is in NEITHER A NOR B . The Venn diagram is as follows:

## Try it

Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.

Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time.

If a student is selected at random, find

• the probability that the student belongs to a club. P ( C ) = 0.40
• the probability that the student works part time. P ( PT ) = 0.50
• the probability that the student belongs to a club AND works part time. P ( C AND PT ) = 0.05
• the probability that the student belongs to a club given that the student works part time.
• the probability that the student belongs to a club OR works part time. P ( C OR PT ) = P ( C ) + P ( PT ) - P ( C AND PT ) = 0.40 + 0.50 - 0.05 = 0.85

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