1. conditional probability

A coin is tossed.
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"?

I proceeded as follows:

Sample space = {(H,H), (H,T), (T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

E = Atleast one tail = {(H,T), (T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

F = the die shows a number greater than 4 = {(T,5),(T,6)}

E intersection F = {(T,5),(T,6)}

P(F/E) = P(E intersection F)/P(E) = (2/8) / (7/8) = 2/7. [n(A intersection B)/n(E)]

They have given the answer as 2/9.

What went wrong I do not know.

Kindly guide me.

Aranga

2. Re: conditional probability

Originally Posted by arangu1508
A coin is tossed.
If the coin shows head, toss it again.
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"
I'm confused with "there is at least one tail".
If the die shows any number, then there MUST have been a tail,
else the die would not be thrown...right?

3. Re: conditional probability

$A=\{\text{die shows > 4}\}$

$B = \{\text{at least 1 tail}\}$

$P[A|B] = \dfrac{P[B|A]P[A]}{P[B]}$

$P[A] = \dfrac 1 2 \dfrac 1 3 = \dfrac 1 6$

$P[B|A] = 1$

$P[B] =1 - P[(H,H)] = 1 - \dfrac 1 4 = \dfrac 3 4$

$P[A|B] = \dfrac{1 \cdot \dfrac 1 6}{\dfrac 3 4} = \dfrac 1 6 \dfrac 4 3 = \dfrac{4}{18} = \dfrac 2 9$

4. Re: conditional probability

Imagine doing this experiment 300 times. 150 times it comes up heads, 150 times it comes up tails.
The 150 times it comes up heads you toss again. 75 times you get heads again, 75 times you get tails.
The 150 times the first toss comes up tails, you roll a single die. The die will show "4 or less" (1, 2, 3, 4) (4/6)(150)= (2/3)(150)= 100 times and will show "greater than 4" (5, 6) (2/6)(150)= (1/3)(150)= 50 times.

"the die shows a number greater than 4" given that "there is at least one tail". "At least one tail" is peculiar wording! The only way you could toss the coin again is if the first toss was a head so it isn't possible to get more than one tail in this experiment. Saying "at least one tail" or "the first toss was a tail" puts us into the second 150 above. In that case, the die roll was greater than 4 2/3 of the time, 100. The probability is 100/150= 2/3.

5. Re: conditional probability

Originally Posted by romsek
$A=\{\text{die shows > 4}\}$

$B = \{\text{at least 1 tail}\}$

$P[A|B] = \dfrac{P[B|A]P[A]}{P[B]}$

$P[A] = \dfrac 1 2 \dfrac 1 3 = \dfrac 1 6$

$P[B|A] = 1$

$P[B] =1 - P[(H,H)] = 1 - \dfrac 1 4 = \dfrac 3 4$

$P[A|B] = \dfrac{1 \cdot \dfrac 1 6}{\dfrac 3 4} = \dfrac 1 6 \dfrac 4 3 = \dfrac{4}{18} = \dfrac 2 9$
sim verifies these numbers