Question

A fair coin is tossed three times,
Ω={hhh,hht,hth,htt,thh,tth,tht,ttt}
X=total number of heads, Y= total number of tails
if the probability of head is 1/2 then
P{hhh} = ?, P{hht,hth,thh}=? how do we calculate these probabilities?

Answer 1

count

Answer 2

p{hhh} = 1/2 + 1/2 + 1/2 =3/2

Answer 3

is it?

Answer 4

out of 8 equally likely outcomes, only one is \(\{h,h,h\}\) so the probability is
\(\frac{1}{8}\)

Answer 5

ok lets go slow
first of all a probability is a number between 0 and 1 inclusive, so never return an answer like \(\frac{3}{2}\)

Answer 6

if you want to compute in this case it is a matter of counting. one case out of the possible 8, and since each outcome is equally likely the answer is \(\frac{1}{8}\)

Answer 7

if you prefer to do a computation, since the probability of getting a "head" on any one toss is \(\frac{1}{2}\) to probability of getting a head on all three tosses is \(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=(\frac{1}{2})^3=\frac{1}{8}\)

Answer 8

ok now lets say that the probability of head on each toss is 2/3 & the probability of tail on each toss is 1/3 then how do we calculate the probability of P{hhh}=?

Answer 9

in other words it is a multiplication and not an addition

Answer 10

but the text says that P = ∑pi

Answer 11

same as before, but now since the probability of getting head on one toss is \(\frac{2}{3}\) the probability of getting it three times in a row is \((\frac{2}{3})^3=\frac{8}{27}\)

Answer 12

& p{hht}=?

Answer 13

ok but presumably \(p_i\) means the probability of one specific outcome.

Answer 14

again there is only one of these
\(\{h,h,t\}\)

Answer 15

i think i see where the confusion is
suppose you were asked "what is the probability of getting 2 heads in 3 tosses?"
then you would have 3 possible outcomes favorable to this event, namely
\(\{h, h, t\},\{h,t,h\}, \{t, h, h\}\) and now you would add the probability associated to each

Answer 16

if they were each equally likely, then the probability of each one is \(\frac{1}{8}\) and so you would get \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\)

Answer 17

but \(\{h,h,t\}\) is only one outcome, so the probability of getting that exactly is \(\frac{1}{8}\)

Answer 18

if the prob of head is 2/3 & tail's 1/3 then what is P{htt}=? & P{hht}=?

Answer 19

sorry wait

Answer 20

in these cases you have to multiply as well

Answer 21

X(hhh)= 3, X(hht)=X(hth)=X(thh)=2, X(htt)=X(tht)=X(tth)=1, X(ttt)=0 where X=total number of heads then if the probability of head on each toss is 2/3 & tail has 1/3 probability then how do we calculate P{X=0}, P{X=1}, P{X=2}, P{X=3} ??????

Answer 22

@satellite73 please help me

Answer 23

ok we can do this quickly because i have to run

Answer 24

ok no problem

Answer 25

in each case multiply
so for example \(\{h,h,t\}\) has probability \(\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\)

Answer 26

but in text it says P{X=2} = 12/27

Answer 27

yes i know that

Answer 28

ok got it

Answer 29

thanks a tonne

Answer 30

we computed for one specific case right?

Answer 31

after the multiplication the summation comes in action

Answer 32

but there are three possibilities, \(\{h,h,t\}, \{h,t,h\}, \{t,h,h\}\)

Answer 33

right, and since each of these outcomes has the same probability, namely
\[\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\] the addition is actually a multiplication, i.e.
\[3\times (\frac{2}{3})^2\times \frac{1}{3}\]

Answer 34

hope this helps i have to run