Question 1

A)
Use the table to answer the question.

Various Outcomes of Different Types of Events
Outcome 1 Outcome 2 Outcome 3 TOTAL
Outcome X 4 9 2 15
Outcome Y 6 4 5 15
TOTAL 10 13 7 30
The table shows the various outcomes of two different events. In order to find the probability of Outcome Y given that Outcome 2 occurs, what number would you use in the denominator of your fraction when using the formula P(A|B)=n(A and B)n(B)
?

(1 point)
Responses

15
15

30
30

10
10

13
13
Question 2
A)
Use the image to answer the question.

A tree diagram splits into 2 levels.

Newlyweds Katherine and Matthias plan to have two children. Assuming it is equally likely to be a boy or a girl, use the tree diagram to help you find the probability that their second child is a girl, given that their first child is a boy.

(1 point)
Responses

12
Start Fraction 1 over 2 End Fraction

18
Start Fraction 1 over 8 End Fraction

34
Start Fraction 3 over 4 End Fraction

14
Start Fraction 1 over 4 End Fraction
Question 3
A)
Use the image to answer the question.

A Venn diagram shows two intersecting circles. Circle 1 represents event A, 3 and 5. Circle 2 represents event B, 6, 8, 10, and 12. The intersection of both the circles is labeled 7, 9, and 11. The space outside the circles is labeled 2 and 4.

The Venn diagram represents the results of the experiment of rolling two fair, six-sided number cubes and adding the sum of the two rolls. Consider event A
as “sum is odd” and event B
as “sum is 6 or greater.” Find the probability that the sum is odd, given that the sum is 6 or greater.

(1 point)
Responses

37
Start Fraction 3 over 7 End Fraction

512
Start Fraction 5 over 12 End Fraction

35
Start Fraction 3 over 5 End Fraction

312
Start Fraction 3 over 12 End Fraction
Question 4
A)How do you determine if the outcome of two events is independent?(1 point)
Responses

Show that P(A and B)=P(A)+P(B)
.
Show that upper P left parenthesis upper A and upper B right parenthesis equals upper P left parenthesis upper A right parenthesis plus upper P left parenthesis upper B right parenthesis .

Show that P(A and B)=P(B)P(A)
.
Show that upper P left parenthesis upper A and upper B right parenthesis equals Start Fraction upper P left parenthesis upper B right parenthesis over upper P left parenthesis upper A right parenthesis End Fraction .

Show that P(A and B)=P(A)⋅P(B)
.
Show that upper P left parenthesis upper A and upper B right parenthesis equals upper P left parenthesis upper A right parenthesis times upper P left parenthesis upper B right parenthesis .

Show that P(A and B)=P(A)P(B)
.
Show that upper P left parenthesis upper A and upper B right parenthesis equals Start Fraction upper P left parenthesis upper A right parenthesis over upper P left parenthesis upper B right parenthesis End Fraction .
Question 5
A)
Use the table to answer the question.

Circle and Triangle Categorized into Different Colors
Color
Shape Red Yellow TOTAL
Circle 12 18 30
Triangle 20 30 50
Total 32 48 80
If event A
is that a shape is a triangle and event B
is that a shape is yellow, are the two events independent? Explain.

(1 point)
Responses

Yes, because P(A)=P(B)
.
Yes, because upper P left parenthesis upper A right parenthesis equals upper P left parenthesis upper B right parenthesis .

Yes, because P(A|B)=P(A)
.
Yes, because upper P left parenthesis upper A vertical bar upper B right parenthesis equals upper P left parenthesis upper A right parenthesis .

No, because both events do not include triangles.

1 answer

Let's answer each question one by one.

Question 1

In order to find the probability of Outcome Y given that Outcome 2 occurs, we need to look at the events defined in the table. The denominator of the fraction we use in the formula P(A|B) = n(A and B) / n(B) is the total number of occurrences of Outcome 2.

From the table:

  • Outcome 2 occurs 9 times for Outcome X and 4 times for Outcome Y.
  • Therefore, the total occurrences for Outcome 2 is 9 + 4 = 13.

The answer is: 13


Question 2

The problem asks for the probability that their second child is a girl, given that their first child is a boy. In a tree diagram illustrating the combinations of children's genders, the possible scenarios given that the first child is a boy (B) are:

  1. B (first), B (second)
  2. B (first), G (second)

Out of these scenarios, only one scenario (B, G) has a girl as the second child. Therefore, the probability that the second child is a girl (given the first child is a boy) is:

P(G | B) = Number of favorable outcomes for G / Total outcomes given first child is B = 1 / 2.

The answer is: 1/2


Question 3

Based on the Venn diagram where:

  • Event A (sum is odd): Odd sums from rolling two dice (1+1, 1+3, etc.)
  • Event B (sum is 6 or greater): Possible sums from 6 to 12.

We need to find the probability that a sum is odd given that the sum is 6 or greater (P(A | B)).

The odd sums that are 6 or greater from sums (7, 9, 11) occur. The odd sums in our region of interest (sums 6 or greater) are 7 and 9.

  • Total sums that are 6 or greater: 6, 7, 8, 9, 10, 11, 12 gives us 7 possible outcomes: {6, 7, 8, 9, 10, 11, 12}
  • Favorable outcomes from this set that are odd (sums {7, 9, 11}):

Thus, the probability P(A|B) = (Number of favorable outcomes for A) / (Total outcomes for B) = 3 / 7.

The answer is: 3/7


Question 4

The correct way to determine if the outcome of two events A and B is independent is:

Show that P(A and B) = P(A) * P(B).

This means that the occurrence of one event does not affect the occurrence of the other event.

The answer is: Show that P(A and B) = P(A) * P(B).


Question 5

To determine if the two events A (shape is a triangle) and B (shape is yellow) are independent, we can check the formula for independence:

P(A and B) = P(A) * P(B).

  • Total triangles = 50, yellow triangles = 30 (P(A) = 50/80; P(B) = 48/80)
  • Calculate P(A and B) = P(yellow triangle) = Number of yellow triangles out of total shapes = 30/80.

Now check:

  • P(A) * P(B) = (50/80) * (48/80) = 2400/6400 = 3/8 (not equal to 30/80)

Since P(Yellow Triangle) is not equal to P(A) * P(B), the events are not independent.

The answer is: No, because P(A and B) ≠ P(A) * P(B).