Question

Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’

Answer 1

diya but there must b number of throws through which we can evaluate

Answer 2

(H,H)(H,T)(T,1)(T,2)(T,3)(T,4)(T,5)(T,6)

Answer 3

yehh now it is correct

Answer 4

:)

Answer 5

hm now?

Answer 6

So we can toss the coin up to six times until we get tail?

Answer 7

atleast one tail probablity= .7
i guess answer is (2/6)/7/10

Answer 8

If the coin shows head toss it again
if it shows tail, then throw a die

Answer 9

thats not the answer

Answer 10

wut the anser

Answer 11

doin it afta 2 year so mayb i did small mistake

Answer 12

Answer is 2/9

Answer 13

I have the solution.. it says
Probabilities assiggned to each event
(H,H)(H,T)(T,1)(T,2)(T,3)(T,4)(T,5)(T,6) are 1/4 ,1/4 ,1/12,1/12,1/12,1/12,1/12,1/12 respectively
idk how?

Answer 14

i m almost done w8

Answer 15

ok

Answer 16

I don't think I understand the answer, but I do see where the numbers come from for
(H,H)(H,T)(T,1)(T,2)(T,3)(T,4)(T,5)(T,6)
do you need help understanding that?

Answer 17

Hmm Yes

Answer 18

http://openstudy.com/#/updates/4efe3765e4b01ad20b52feee
when this is done thx

Answer 19

probabilites r assigned in a way like (H,H) it means prob of getting h is .5 then next prob is also .5 so we multiply them to get 1/4 for this bracket.
for a particular number, the probability of getting tales it .5, and then gettin any number suppose 5, is 1/6, so probabilty (T,5) is 1/6*1/2=1/12

Answer 20

I make it:\[p=\frac{1}{3}(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)=\frac{1}{3}*1=\frac{1}{3}\]