Tree probability

Question

I m posting one question without options. I want to know the multiple way to solve the question without actually counting the events.

Q.1 A coin is tossed. If Head, then a die is thrown. If tail, the coin is tossed again.

Find the probability of:
1. Two tails.
2. Head and Number 6.
3. Head and an even number.

Since it is a small problem, we can actually solve by counting the pairs. But I want to solve it using probability tree. I tried it using the one stated in Math Book, but the different answer came.

shrouded1 · Answer

Two tails = 0.5 * 0.5 = 1/4
heads + 6 = 0.5 * 1/6 = 1/12
heads + even = 0.5 * 0.5 = 1/4

FQ · Answer

the best "tree" I could do in text:

--- (1/2) --- head, use die --- (1/6) --- roll of 1 *path 1a
--- (1/2) --- head, use die --- (1/6) --- roll of 2 *path 1b
--- (1/2) --- head, use die --- (1/6) --- roll of 3 *path 1c
--- (1/2) --- head, use die --- (1/6) --- roll of 4 *path 1d
--- (1/2) --- head, use die --- (1/6) --- roll of 5 *path 1e
--- (1/2) --- head, use die --- (1/6) --- roll of 6 *path 1f

--- (1/2) --- tail, flip coin --- (1/2) --- head *path 2a
--- (1/2) --- tail, flip coin --- (1/2) --- tail *path 2b

(a) two tail = path 2b = (1/2)*(1/2) = 1/4
(b) heads, then roll of 6 = path 1f = (1/2)*(1/6) = 1/12
(c) heads, then roll of even # = path 1b + path 1d + path 1f
 = (1/2)*(1/6) + (1/2)*(1/6) + (1/2)*(1/6) = 1/4
 you add them up becoz they are independent events

gurpreetsingh · Answer

I did the same way and got the same answer. It is not correct :-D

Try again.

shrouded1 · Answer

Just to be clear about the question, there are exactly 2 tosses right ?

If you toss a tail, re-toss and get a heads, do you need to roll the dice ?

gurpreetsingh · Answer

I have posted the exact question. I m not sure about the intent. I got exactly the same answer.

THE PROBABILITY OF OBTAINING : -> this is the statement.

In the solution, all the pairs have been counted and then probability is taken.

shrouded1 · Answer

The only other thing I can think of is if the question means that you keep tossing till you get a heads and then you do a dice throw. In which case :

P(T,T) = 1/4
P(H,6) = 1/6*(1/2+1/4+1/8+.....) = 1/6
P(H,Even) = 1/2*(1/2+1/4+1/8+...) = 1/2

gurpreetsingh · Answer

I think in the question we just have two conditions. Either A or B. So the total outcomes will be sum of both the events. If we manually count the pair of events then probability is different from what you have got. Bunnel: We need your help !!

shrouded1 · Answer

You have really sparked my curiosity.
I hope you dont mean counting {H1,H2,H3,H4,H5,H6,TT,TH}
Because each of those outcomes is not equiprobable

gurpreetsingh · Answer

Exactly !!!

I got the same answer you got earlier in the post, but the explanation used the above sample space.
This question is from 11th RD sharma text book. I believe the book is authentic.

He has solved the question using the above sample space. My question is why the above method is correct/incorrect?

shrouded1 · Answer

Its def not correct. It can't be, not for an unbiased coin and an unbiased dice.

As for the authenticity of R.D.Sharma - well, clearly we have evidence here otherwise ! You can't count and solve because events are not equi-probable so you cannot reduce this to a set selection problem

Tree probability

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