Question

probability: determining probabilities
ii.a student activity sheet 2: using tree diagrams
as president of the high school band, catrina needs to pick a committee of 2 to accompany
her each time she visit middle schools. the director told her that each committee had to
consist of 1 boy and 1 girl: 5 boys and 4 girls volunteer to go. to be fair, catrina makes a
spinner with the boys names and a spinner with the girls names. each time she schedules a
visit, catrina spins each spinner once to determine who goes with her. if a spinner lands on a
line, she spins again.

1. draw a tree diagram to show all the possible combinations of volunteers who might go

with catrina. how many outcomes are in the sample space?

1. are all the outcomes equally likely? what would make the outcomes not equally likely?
2. what is the probability that nathan will be selected? explain your reasoning. list the

possible outcomes for 2 - person committees that include nathan.

1. if ave decides she cannot go on a visit she is scheduled for, how does this change the

probability for other boys or other girls to be selected? explain your reasoning.

Explanation:

Step1: Determine the number of boys and girls

There are 5 boys (Nathan, Spence, Zenil, Will, Xavier) and 4 girls (Ave, Lauren, Pam, Deb).

Step2: Calculate the total number of possible combinations

For each boy, there are 4 possible girls. So the total number of combinations in the sample space is \(5\times4 = 20\)

Step3: Check if outcomes are equally likely

Since each spinner is fair (lands on each name with equal probability), the probability of each boy - girl combination is \(\frac{1}{5}\times\frac{1}{4}=\frac{1}{20}\). So the outcomes are equally likely.

Step4: Calculate the probability that Nathan is selected

The number of combinations that include Nathan is 4 (Nathan - Ave, Nathan - Lauren, Nathan - Pam, Nathan - Deb). The probability \(P=\frac{4}{20}=\frac{1}{5}\)

Step5: Analyze the effect of Ave's decision

If Ave can't go, the number of girls becomes 3. For each boy, there are now 3 possible girls. The total number of combinations is \(5\times3 = 15\). For boys: the probability of a boy being selected is still \(\frac{1}{5}\) (because the number of boys hasn't changed). For girls (ex - Ave): the probability of each of the remaining 3 girls is \(\frac{1}{3}\) (for a particular girl) given a boy is selected. The overall probability for a girl (e.g., Lauren) is \(\frac{1}{5}\times\frac{1}{3}=\frac{1}{15}\) (previously \(\frac{1}{20}\)). So the probability for the remaining girls increases.

Answer:

1. There are 20 possible combinations in the sample space.
2. Yes, the outcomes are equally likely because each spinner is fair.
3. The probability that Nathan is selected is \(\frac{1}{5}\). The possible outcomes with Nathan are (Nathan, Ave), (Nathan, Lauren), (Nathan, Pam), (Nathan, Deb).
4. The probability for boys remains \(\frac{1}{5}\) (number of boys unchanged). The probability for the remaining girls (Lauren, Pam, Deb) increases. Previously, for a girl, the probability was \(\frac{1}{20}\) (e.g., \(P(\text{Lauren})=\frac{1}{5}\times\frac{1}{4}\)), now \(P(\text{Lauren})=\frac{1}{5}\times\frac{1}{3}=\frac{1}{15}\)