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?
Link:Page 7 http://ncertbooks.prashanthellina.com/class_12.Mathematics.MathematicsPartII/Probability%2018.11.06.pdf
Best Answer
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 $$\frac {\frac 2{12}}{\frac 34}=\frac 2{12}\times \frac 43 =\frac 8{36}=\frac 29$$