Math Is Fun Forum

juantheron

Member

Registered: 2011-10-19

Posts: 312

dice problem

consider the experiment of tossing a coin . if the coin shows head toss it again but if it shows tail then throw a dice then

find the conditional  probability of the event 'the dice shows a no.  greater then 4'. given that there is at least one tail

my solution ::

sample space S={HH,HT,T1,T2,T3,T4,T5,T6}

Let B= event dice shows >4

and A=at least one tail

So

but answer given is

So where i have done wrong

can anyone explain me

thanks

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: dice problem

Hi;

The second answer is right although your notation is slightly off.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: dice problem

yes bobbym you are right 2nd one is right

i have a question why 1 st answer is not right

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: dice problem

Hi;

Please see post #2.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: dice problem

yes how can i get it (post 2)

and where i have done wrong

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: dice problem

Hi;

B= event dice shows >4

A=at least one tail

The probability of A and B both happening is

(1/2)(1/3) = 1/6 that is the numerator.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: dice problem

Thanks bobbym (you are right)

sample space S={HH,HT,T1,T2,T3,T4,T5,T6}

Let B= event dice shows >4

and A=at least one tail

here n(B)= {T5,T6} and n(A)={HT,T1,T2,T3,T4,T5,T6}

So

So i have a doubt that why this is not a answer

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: dice problem

You are counting the sample space as if they are all equally probable.

Follow the formula:

I explained how to get the numerator above it is 1/6. The probability of A is 1/2 + 1/4. Why? Because you can get a tail on the first or the second throw. When you see or that means add.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: dice problem

yaa i partially understand

i have last doubt

what is the meaning of equaly probable here

thank

Last edited by juantheron (2011-12-16 07:08:41)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: dice problem

HT is 1/4
T4 is 1/12

They are not equally probable therefore you can not add them up like you would add (1,2,3) up.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: dice problem

yes bobbym fully  got it

thanks

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: dice problem

Hi;

Are you sure because I will provide more if you need it?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

juantheron

Member

Registered: 2011-10-19

Posts: 312

Re: dice problem

no no got it