How do you solve these probability problems from the SAT?
December 20th, 2009 | 
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Hi and welcome to my blog,
I opened this blog to help anyone who is preparing for his SAT.
I remember myself when I started - I was so lost and in such a panic.
Luckily I got this great SAT guide that helped me a lot . So study hard and good luck!
My SAT tutor has a rule that when a probability problem involves "AND" you multiply, and when a probability problem involves "OR" you add. I wasn’t there for that class but I have the answers to the problems. I just don’t know how to get them. If you answer please show all steps of how you got the answer using the method my tutor uses.
Thanks!
Norma tosses 5 Mounds, 7 Almond Joys, and 4 Snickers into a bag.
1. What is the probability that Seymour will grab three Snickers and 1 mounds?
2. If he hates Snickers and will discard any Snickers he gets, but he will eat anything else, what’s the probability that the first bar he both grabs and eats will be a Mounds?
What about if you do these with or?
Thanks!
Actually Murf the Surf, I think you did something wrong there. The answers to these questions are:
1. 1/364
2. 5/12
Any other ideas on how to do this?
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let us use the initial letters to identify the bars: M , A , S
q1
P[SSSM] = [4/16][3/15][2/14][5/13] =120/43680 = 1/364
[the denominators go on reducing because
you aren't replacing the bars]
note:
——
if 1/364 is the correct answer, the question shd hv
SPECIFIED one particular order, e.g. SSSM.
if it is not specified, the ans shd be multiplied by 4!/3! or 4
take that up with your teacher !
q2
the discarded snickers don’t count,
he has to grab a M before an A
so P[M] = 5/12
For question #1, the probability that Seymour (reaching into the bag and grabbing 4 candy bars) will get 3 Snickers and one Mounds is: 4/16 x 4/16 x 4/16 x 5/16 or, when reduced, 5/1024.
#2 There are 16 bars total and 5 of them are mounds so the probability of getting a Mounds is 5/16 on his first grab.
Q. #1 you can think of as reading: "What is the probability that Seymour will grab a Snicker AND another Snicker AND another Snicker AND also a Mounds?" You multiply the probability of getting each individual bar together to get the total probability of getting just three Snickers and one Mounds all at once.
The "or" part would be like: what is the probability of Seymour reaching in and getting either a Mounds OR a Snickers? In that case you would add the two probabilities together: 5/16 + 4/16 which would be 9/16 (more than likely).