Probability

Advanced Math 7 · Unit 16 · probability-trees · Teacher edition

probability-trees

Name ______________________________Date ______________

1.
A bag holds $3$ red, $2$ blue, and $5$ green marbles. One marble is drawn at random. What is the probability that it is not green? Give a fraction in lowest terms.
Answer: ______________
2.
A bag holds $4$ red and $2$ blue counters. Two counters are drawn one after the other without replacement. What is the probability that both are the same colour? Give a fraction in lowest terms.
Answer: ______________
3.
Two fair six-sided dice are rolled. What is the probability that the two numbers add to $8$ ? Give a fraction in lowest terms.
Answer: ______________
4.
Two fair coins are tossed, one after the other. What is the probability of getting heads on the first coin and tails on the second?
1. (A)
  $\dfrac{1}{4}$
2. (B)
  $\dfrac{1}{2}$
3. (C)
  $\dfrac{3}{4}$
4. (D)
  $1$
5.
A bag holds $2$ red and $3$ blue marbles. Two marbles are drawn at random without replacement. What is the probability that at least one of them is red?
1. (A)
  $\dfrac{3}{10}$
2. (B)
  $\dfrac{2}{5}$
3. (C)
  $\dfrac{3}{5}$
4. (D)
  $\dfrac{7}{10}$
5. (E)
  $\dfrac{4}{5}$

Answer key — Probability

1.
1/2probability-trees-01
$P(\text{not green}) = \frac{3 + 2}{10} = \frac{5}{10} = \mathbf{\tfrac{1}{2}}$ .
2.
7/15probability-trees-02
$P(RR) = \frac{4}{6} \cdot \frac{3}{5} = \frac{12}{30}$ ; $P(BB) = \frac{2}{6} \cdot \frac{1}{5} = \frac{2}{30}$ . Sum $= \frac{14}{30} = \mathbf{\tfrac{7}{15}}$ . (With replacement the answer would be $\frac{16}{36} + \frac{4}{36} = \frac{5}{9}$ .)
3.
5/36probability-trees-03
Five of the $36$ cells have sum $8$ : $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ . $P = \mathbf{\tfrac{5}{36}}$ .
4.
(A)
$\dfrac{1}{4}$
probability-trees-04
$P(H \text{ then } T) = \frac{1}{2} \times \frac{1}{2} = \mathbf{\tfrac{1}{4}}$ — one of the four equally likely outcomes $HH, HT, TH, TT$ . ( $\frac{1}{2}$ counts only one coin; $1$ comes from adding $\frac{1}{2} + \frac{1}{2}$ along the branch.)
5.
(D)
$\dfrac{7}{10}$
probability-trees-05
$P(\text{both blue}) = \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}$ , so $P(\text{at least one red}) = 1 - \frac{3}{10} = \mathbf{\tfrac{7}{10}}$ . ( $\frac{3}{10}$ is the probability of no red; $\frac{2}{5}$ is only the chance that the first marble is red.)