# Probability of getting number greater than 4 given there is at least one tail .(Conditional Probability)

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’.

Now,I know how to do this question using multiplication theorems of probability and the answer is 2/9 but the book solves this question before starting the topic of multiplication theorems.

So my question is simple?Can it be solved without using multiplication theorems and using the basic probability knowledge,preferably after knowing addition theorem and conditional probability but before multiplication theorem?Any hints?

• Have you reproduced the exact question ? Feb 23, 2016 at 11:20
• @true blue anil Yes.The 1st paragraph.After that my question. Feb 23, 2016 at 11:22
• You only do one round of this? So, given that there is "at least one tail" means "the single toss of the coin comes up tails"? But then the answer is $\frac 26$.
– lulu
Feb 23, 2016 at 11:26
• @lulu Yes,only one round,I have added the link also. Feb 23, 2016 at 11:28
• Oh, I get it. I'll post something below.
– lulu
Feb 23, 2016 at 11:36

I think the problem is (somewhat) ambiguous. As I (now) read it the coin portion of the game has three possible outcomes $$\{HH,HT,T\}$$ These are not equally likely. Indeed the first two have probability $\frac 14$ each, and the last has probability $\frac 12$. Only in the latter event do you actually roll the die (thus I think the problem ought to read "given that you the coin ends up $T$ what is the probability that you throw the die and see a value greater than $4$." ) We note that the probability of the coin portion ending in $T$ is $\frac 34$

Anyway, now you can do a tree. The outcomes that involve a $T$ are $\{T,_\}$ with probability $\frac 14$ and $\{T,i\}$ for $i\in\{1,2,3,4,5,6\}$ and each of these has probability $\frac 1{12}$. We note that exactly two of these events pass the "greater than $4$" test and that these events have a combined probability of $\frac 2{12}$. Thus our answer is $$\frac {\frac 2{12}}{\frac 34}=\frac 2{12}\times \frac 43 =\frac 8{36}=\frac 29$$

• So how did you get 1/4 without multiplying 1/2 with itself,I mean how did you get 1/4 using only the sample space?(Because I want to know how to get probability in this question without using multiplication theorem) Feb 23, 2016 at 11:47
• I did multiply them. That's exactly how it is done in your book. They draw the tree and carefully write out the probability along each branch (by multiplying). My argument isn't all that different, I think it's just a little cleaner.
– lulu
Feb 23, 2016 at 11:50
• But ,but they taught the topic after doing this question,anyway your answer is helpful,that must be incorrect placement in the book.Thanks:) Feb 23, 2016 at 11:52
• Well, if you really want to avoid the multiplication...enumerate the coin part as $4$ equi-probable states $\{HH,HT,TH,TT\}$ Here the last two states are summed up as what I called just $T$ before. Now it's clear that $HT$ has probability $\frac 14$.
– lulu
Feb 23, 2016 at 11:56
• And for the tails and dice I would consider it(sample space) for both Heads and tails and 6 numbers of the dice to get 12 outcomes or something else? Feb 23, 2016 at 11:59