Help? Please! A math class has 3 girls and 7 boys in the seventh grade and 6 girls and 2 boys in the eighth grade. The teacher randomly selects a seventh grader and an eighth grader from the class for a competition. What is the probability that the students she selects are both girls? Write your answer as a fraction in simplest form.
The events 'select a seventh grader' and 'select an eigth grader' are independent. The outcome of one has absolutely no effect on the other. In this case the probability of both occurring is the product of their probabilities: \[P(A \cap B)=P(A)\times P(B)\] There are 10 students in the seventh grade, 3 of which are girls. Can you find the probability of randomly selecting a girl?
which numbers would I use as the variables? Once I have those I think I could find the answer.
The numbers to work out the probability are not variables. I asked you 'There are 10 students in the seventh grade, 3 of which are girls. Can you find the probability of randomly selecting a girl?' The numbers that you need are in this question.
@maddiemaze Are you there?
Yes I'm sorry I'm still not sure how to do this
'There are 10 students in the seventh grade, 3 of which are girls. Can you find the probability of randomly selecting a girl?' Probability of randomly selecting a girl is \[P(girl)=\frac{number\ of\ girls}{number\ of\ students}=\frac{3}{10}\] Can you use this example to find the probability of randomly selecting a girl from the eighth grade where there are 8 students, 6 of which are girls?
6/8?
Correct! Well done. Now the final stage is to find the required probability by multiplying the two values of probability as follows: \[P(2\ girls)=\frac{3}{10}\times \frac{6}{8}=you\ can\ calculate\] Hint: Remember to simplify your answer.
\[\frac{3}{10}\times \frac{6}{8}=\frac{3\times 6}{10\times 8}=?\]
18/80
Correct again! To simplify this fraction divide the numerator by 2 and divide the denominator by 2. Can you do that?
9/40
Correct. You have the required value of probability!
Thanks for your help :)
You're welcome :)
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