Algebra 2 - stats and probability
posted by Kay on .
I need someone to tell me how to set these up please?
1. A box contains 4 quarters. 2 of the quarters have the American eagle on the back. Suppose you draw 2 quarters at random with replacement.
a. What's the probability that both have the American eagle on the back?
b. What's the probability that neither have the American eagle?
c. What's the probability that AT LEAST one has the American eagle?
2. In a survey, 28% of students said they were left-handed. 64% said they were right-handed and 8% said they were ambidextrous. If you omitted the ambidestrous people, what percent of the students remaining are left-handed?
3. 75% of boys like hockey. 65% of girls like hockey. There are an equal number of boys and girls. If someone does not like hockey, what's the probability that it's a boy?
Let E be 'eagle on back"
let N be 'not eagle on back'
There are only 4 possible outcomes
EE, NE, EN, and NN
since there is replacement each of those above events has a prob of 1/4 from (2/4 x 2/4)
so a) is 2/4 x 2/4 = 1/4
b) NN or 1/4 again
c) is 1/4 + 1/4 = 1/2
for conveniece sake, let's say there are 100 boys, then there are 100 girls or a total of 200
the number of boys who don't like hockey is 25
so prob that a boy doesn't like hockey = 25/200 = 1/8
do #2 the same way.
But you're taking the number of boys from the percent of people that don't like hockey. 60% of people don't like hockey. So it's 25%/60%, right? So, wouldn't it be 42%?
That's why I wanted you to pick an arbitrary number, I used 100 because it so convenient.
Can't we break down our 200 students into 4 categories.
boys who like hockey = 75
boys who don't like hockey = 25
girls who like hockey = 65
girls who don't like hockey = 35
(notice that totals 200)
prob(of some event) =number of cases in that event / total number of cases
our event is (boys who don't like hockey)
so prob = 25/200 = 1/8
I don't see how you can interpret that in any other way.