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Mathematics 22 Online
OpenStudy (anonymous):

Need Help fast!!! use the following table to answer the questioin... X = P(X) A = 0.30 B= 0.20 C = ? D = 0.28 Find P(A and D) assuming that A and D are independent.

jimthompson5910 (jim_thompson5910):

P(A and D) = P(A) * P(D) where events A and D are independent

jimthompson5910 (jim_thompson5910):

So just multiply the corresponding probabilities

OpenStudy (anonymous):

0.30 * 0.28 = 0.084

jimthompson5910 (jim_thompson5910):

you nailed it

OpenStudy (anonymous):

Thanks! :)

jimthompson5910 (jim_thompson5910):

yw

OpenStudy (anonymous):

what about this one? you toss 2 coins. find P ( at least 1 tail is tossed)

jimthompson5910 (jim_thompson5910):

What's the sample space look like? Any ideas?

OpenStudy (anonymous):

TT TH HT HH

jimthompson5910 (jim_thompson5910):

good

jimthompson5910 (jim_thompson5910):

How many have at least one tail? In other words, how many have one T or two Ts?

OpenStudy (anonymous):

3

jimthompson5910 (jim_thompson5910):

out of how many?

OpenStudy (anonymous):

4

jimthompson5910 (jim_thompson5910):

Good, so P ( at least 1 tail is tossed) = 3/4

OpenStudy (anonymous):

but since their are 2 coins, wouldn't it make it 6/8?

jimthompson5910 (jim_thompson5910):

no, there are only 4 possible ways to toss 2 coins

jimthompson5910 (jim_thompson5910):

and there are only 3 ways to get at least one tail

jimthompson5910 (jim_thompson5910):

besides, 6/8 reduces to 3/4 anyway

OpenStudy (anonymous):

oh okay :) for some reason, it keeps giving me the answer, 7/8??

jimthompson5910 (jim_thompson5910):

really? hmm odd

jimthompson5910 (jim_thompson5910):

are you sure that it isn't 3 coins?

jimthompson5910 (jim_thompson5910):

if you toss 3 coins, then P(at least one tail) = 7/8

OpenStudy (anonymous):

ohhhh, I see now :P

jimthompson5910 (jim_thompson5910):

this is because the total possible ways to toss 3 coins are HHH HHT HTH HTT THH THT TTH TTT and 7 of these total 8 have at least one tail.

OpenStudy (anonymous):

ahhh, okay :) hmmm what about this? When tossing 2 coins, P ( at least 1 head) = P(not all tails) True or False?

jimthompson5910 (jim_thompson5910):

Well we know that P ( at least 1 head) = 3/4 from last time right?

OpenStudy (anonymous):

yes

jimthompson5910 (jim_thompson5910):

What about P(not all tails)?

OpenStudy (anonymous):

= 1/4

jimthompson5910 (jim_thompson5910):

what is P(all tails)?

OpenStudy (anonymous):

0/4?

jimthompson5910 (jim_thompson5910):

Remember, we have HH HT TH TT How many of the 4 are all tails?

OpenStudy (anonymous):

4

jimthompson5910 (jim_thompson5910):

HH has all tails?

OpenStudy (anonymous):

so 3?

jimthompson5910 (jim_thompson5910):

HT is all tails?

OpenStudy (anonymous):

1?

jimthompson5910 (jim_thompson5910):

good, only TT is all tails

jimthompson5910 (jim_thompson5910):

so P(all tails) = 1/4

jimthompson5910 (jim_thompson5910):

Now we can use this idea to find P(not all tails)

OpenStudy (anonymous):

2/4

jimthompson5910 (jim_thompson5910):

close, but not quite

jimthompson5910 (jim_thompson5910):

P(not all tails) = 1 - P(all tails)

OpenStudy (anonymous):

0?

jimthompson5910 (jim_thompson5910):

P(not all tails) = 1 - P(all tails) P(not all tails) = 1 - 1/4 P(not all tails) = ???

OpenStudy (anonymous):

0.75

jimthompson5910 (jim_thompson5910):

or 3/4

jimthompson5910 (jim_thompson5910):

so because P(not all tails) = 3/4 and P ( at least 1 head) = 3/4, we can say that P ( at least 1 head) = P(not all tails)

OpenStudy (anonymous):

So that answer is indeed True

jimthompson5910 (jim_thompson5910):

you got it

OpenStudy (anonymous):

Thanks for your help! Really appreciate it! :)

jimthompson5910 (jim_thompson5910):

you're welcome

OpenStudy (anonymous):

I have trouble figuring these types of questions out, could you help? his is an example of which of the following sampling techniques? HELP? A chef wants to open a new Italian restaurant. He wishes to find out which kinds of pasta (spaghetti, linguini, or penne) will be the most popular with potential customers. He hires a polling organization to survey the market. The company builds a telephone survey list by selecting one person at random from the first page of the telephone book, who turns out to be the 123rd person on the page. The company then fills out the list by adding the 123rd person on every page of the book. A.Systematic sample B.Stratified sample C.Simple random sample (SRS) D.Cluster sample E. Convenience sample

jimthompson5910 (jim_thompson5910):

Any ideas?

OpenStudy (anonymous):

I'm thinking its systematic because, hes choosing the 123rd person, and then 123td person on every page of the book?

jimthompson5910 (jim_thompson5910):

you got it, very nice

OpenStudy (anonymous):

thanks !

jimthompson5910 (jim_thompson5910):

yw

OpenStudy (anonymous):

How do you find the mean, using a histogram?

jimthompson5910 (jim_thompson5910):

each value under the histogram represents a value in the data set the heights of each histogram represent how frequent each value in the data set occurs making sense so far?

OpenStudy (anonymous):

Yes, and it also says this histogram as a class width of 1

jimthompson5910 (jim_thompson5910):

So you simply multiply each value under the histogram by the corresponding height...then you add up each product Finally, you divide that last result by the total number of values (ie the sum of the frequencies)

OpenStudy (anonymous):

Okay, so the Height is 5..... and the width is 10 but there are only 9 being used. so 1 * 2 3 * 4 4 * 5 5 * 4 6 * 2 7 * 1 8 * 1 9 * 1 10 * 1

jimthompson5910 (jim_thompson5910):

Multiply all those out, add up the results and divide that by the sum of the heights

OpenStudy (anonymous):

632 / 15 Is this right?

jimthompson5910 (jim_thompson5910):

hmm I'm getting something different, but that could mean that I'm missing something (since I can't see the histogram)

OpenStudy (anonymous):

well, I'll see if I can post it here... but basically it has a height of 5, and a width of 10 but only nine blocks are used.

jimthompson5910 (jim_thompson5910):

which one has a height of 5?

OpenStudy (anonymous):

|dw:1341175385563:dw|

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