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Alma Said:A couple high school level math problems?
We Answered:This equation is invalid, the left and right sides are not equal so therefore there is no solution.
It might be an error in the typing.
Stephen Said:high school math: probability. please help!?
We Answered:* Four coins are tossed. Find the probabitily that AT LEAST three of the four come up heads.
State the chance of each outcome. For tossed coins, it's a have chance for every head or tail.
List the permutations. For example:
head head head head,
head head head tail,
head head tail head,
head tail head head
...and so on.
The chance of any one of these working out can be found by multiplying.
1/2 * 1/2 * 1/2 * 1/2 ... which is 1/16, so that's the chance of each line in your list of permutations.
Five outcomes satisfy the 'at least three heads' requirement, so the chance is 5/16.
* A penny, a nickel, and a dime are tossed. Specify the event that at most one coin turns up tails.
List the permutations again.
heads heads heads
heads heads tails
heads tails heads
tails heads heads
tails heads tails
tails tails heads
heads tails tails
tails tails tails
Each has a 1 in 8 chance of happening because 1/2 * 1/2 * 1/2 = 1/8
How many satisfy your 'tails' requirement? Four.
So the chance is 4/8 ... which is 1 in 2.
* At a high school there are two administrators and three counselors who park in three reserved places. In how many ways could these spaces be occupied by the cars belonging to these people?
This question has no probability in it, but requires you to list all the different combinations that might happen. Thus:
admin admin counselor
admin counselor admin
counselor admin admin
counselor counselor admin
counselor admin counselor
admin counselor counselor
counselor counselor counselor
There's your answer. There are seven different ways in which the spaces could be used.
* Mr. Rhee's car has a probability of 70% of starting, and Ms. Rhee's car has an 80% probability of starting. What is the probability that
p(Mr. Rhee's car starting) = 0.7
p(Mrs. Rhee's car starting) = 0.8
1. p(neither starts) = 0.3 * 0.2 ... or (1 - 0.7) * (1 - 0.8) = 0.06 ... or 6%
2. p(both start) = 0.7 * 0.8 = 0.56 ... or 56%
3. p(either both cars or neither car will start) =
p(neither starts) + p(both start) = 0.56 + 0.06 ... or 62%
4. p(exactly one of the cars will start) =
p(Mr. Rhee's car starting) * p(Mrs. Rhee's car NOT starting) +
p(Mr. Rhee's car NOT starting) * p(Mrs. Rhee's car starting)
= (0.7 * 0.2) + (0.8 * 0.3) = 0.14 + 0.24 = 0.38 ... or 38%