Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (narusamisty):

Can someone plz help me with this probability problem!! :D Will give medal :) 2. Create a probability distribution table for the following experiment. A coin is tossed two times and after each toss a head or tails is recorded. Let the discrete random variable represent the number of heads in the two tosses.

satellite73 (satellite73):

a coin is tossed twice there are \(2\times 2=4\) equally likely outcomes not such a large number that we can't list them all \[\{HH, HT, TH, TT\}\]

satellite73 (satellite73):

what are the possible number of heads tossed?

OpenStudy (narusamisty):

4 times? :) but im not sure on how to graph it

satellite73 (satellite73):

whoa hold the phone

satellite73 (satellite73):

lets forget about a graph for a second

satellite73 (satellite73):

you got two coins, and not that it makes any difference, lets say the first is a dime, the second a nickel

satellite73 (satellite73):

you toss them both do you understand what i meant by \[\{HH, HT, TH, TT\}\] is the set of possible outcomes ?

OpenStudy (narusamisty):

sorry the HH,HT,TH,TT confused me

satellite73 (satellite73):

ok so lets say the first coin is a dime, the second in a nickel then \(HT\) means the dime was Heads and the nickel was Tails

satellite73 (satellite73):

is it more clear now?

OpenStudy (narusamisty):

nope im still confused :(

satellite73 (satellite73):

ok lets try with a picture

OpenStudy (narusamisty):

okay :)

satellite73 (satellite73):

satellite73 (satellite73):

do you understand that table?

OpenStudy (narusamisty):

yes :)

satellite73 (satellite73):

ok so now you see there are four possible equaly likely outcome right? i just abbreviated them as \[\{HH, HT, TH, TT\}\]

OpenStudy (narusamisty):

so then it would be 3/4 of the possible number of heads tossed or am i still wrong? :)

satellite73 (satellite73):

ok now we got all possible outcomes, now lets see what the possible number of heads are in two rolls

OpenStudy (narusamisty):

okay

satellite73 (satellite73):

start with \(HH\) how many heads?

OpenStudy (narusamisty):

1? :)

satellite73 (satellite73):

???

OpenStudy (narusamisty):

or is it 2? ._.

satellite73 (satellite73):

yes, 2

satellite73 (satellite73):

so one possibility is you get two heads how about \(HT\)?

OpenStudy (narusamisty):

1 right?

satellite73 (satellite73):

right m

satellite73 (satellite73):

and \(TH\)?

OpenStudy (narusamisty):

1 as well?

satellite73 (satellite73):

yes and finally \(TT\)

OpenStudy (narusamisty):

0 :)

satellite73 (satellite73):

yeah good so if we have a random variable, usually denoted by \(X\) there are three possible values for \(X\) \[X=0,X=1,X=2\]

satellite73 (satellite73):

now for the distribution \[X=0\] means you get no heads, i.e. two tails how many ways are there to do that?

OpenStudy (narusamisty):

what do you mean how many are there to do that? :(

satellite73 (satellite73):

there are four possible outcomes , abbreviated as \[\{HH, HT, TH, TT\}\] out of those, how many are TT (it is a silly question kind of, since the answer is obvious)

OpenStudy (narusamisty):

ah it would be 1 :)

satellite73 (satellite73):

right and since there are four equally likely outcomes, \[P(X=0)=\frac{1}{4}\]

satellite73 (satellite73):

now how many ways can you get \[X=2\]

OpenStudy (narusamisty):

3 times right? :)

satellite73 (satellite73):

nope

satellite73 (satellite73):

how may tosses has two heads?

satellite73 (satellite73):

how many HH do you see in \[\{HH, HT, TH, TT\}\]

OpenStudy (narusamisty):

ohh ok i was thinking of the possible heads that i could get

OpenStudy (narusamisty):

it would also be 1

satellite73 (satellite73):

right so \[P(X=2)=\frac{1}{4}\] and finally how many ways can \(X=1\)?

OpenStudy (narusamisty):

does that go for the TH?

OpenStudy (narusamisty):

2? :)

satellite73 (satellite73):

yes so \[P(X=1)=\frac{2}{4}\] which you want to reduce to say \[P(X=1)=\frac{1}{2}\]

OpenStudy (narusamisty):

okay :)

OpenStudy (narusamisty):

now is it that i have to graph it? :)

satellite73 (satellite73):

how all we need to do is write the complete probability distribution which we already have \[P(X=0)=\frac{1}{4}, P(X=1)=\frac{1}{2}, P(x=2)=\frac{1}{4}\]

satellite73 (satellite73):

i don't know what the table is supposed to look like

satellite73 (satellite73):

you got an example?

OpenStudy (narusamisty):

yes wait a sec :)

OpenStudy (narusamisty):

sorry it wont submit is it ok if i draw it

satellite73 (satellite73):

sure

OpenStudy (narusamisty):

|dw:1472697774569:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!