Question

Samantha is playing games at the fair. In the first game, she gets to throw a softball and can score 0, 2, 4, 6, or 8 points. For the second game, she gets to toss a ring and can score 0, 1, 3, or 5 points. The table below shows all of the possible outcomes for playing each game once.
0 1 3 5 OUTCOMES
0 0,0 0,1 0,3 0,5
2 2,0 2,1 2,3 2,5
4 4,0 4,1 4,3 4,5
6 6,0 6,1 6,3 6,5
8 8,0 8,1 8,3 8,5

What is the probability of getting a number greater than 2 in the first game and a number greater than 1 in the second game?
A. 9/20

B. 3/10

C. 1/3

Answer 1

Answer:

          $\large\boxed{\large\boxed{\text{Option B. }3/10}}$

Explanation:

I will rewrite the table for better understanding:

Table

           OUTCOMES

        0        1       3       5

0     0,0     0,1    0,3    0,5

2     2,0     2,1     2,3    2,5

4     4,0     4,1     4,3    4,5

6     6,0     6,1     6,3    6,5

8     8,0     8,1     8,3    8,5

What is the probability of getting a number greater than 2 in the first game and a number greater than 1 in the second game?

The probability of an event is calculated as the quotient of the number of favorable outcomes and the number of total possible outcomes.

$Probability(event)=\frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$

i) Favorable outcomes:

The total number of favorable outcomes is the combination of outcomes with a 4, 6, or 8 in the first number (representing getting a number greater than 2 in the first game) and a 3 or 5 in the second number (representing getting a number greater than 1 in the second game).

Those are: (4,3); (4,5); (6,3); (6,5); (8,3); and (8,5).

Which is a total of 6 favorable outcomes.

ii) Possible outcomes:

That is the total sample space, i.e. 4 × 5 = 20 outcomes.

iii) Compute the probability:

    $Probability=6/20=3/10$