1. ## coin tossed

a coin is tossed 3 times. If both heads and tails appear find the probability that exactly one tail appears.

i am using the conditional probability rule so p(probability of getting one tail | probability of getting one tail and one head )=
p(probability of getting one tail intersection probability of getting one tail and one head)/p(probability of getting one tail and one head)

i am having trouble finding the probability of getting one tail and one head as it could be in any order so i cant say (.5)^3

2. ## Re: coin tossed

Originally Posted by markosheehan
a coin is tossed 3 times. If both heads and tails appear find the probability that exactly one tail appears.
Toss a coin three times. That produces eight elementary outcomes. By the conditional, we eliminate all heads & all tails. Leaving six outcomes. Of those six only three consists of two heads and one tail. Thus what is the probability?

3. ## Re: coin tossed

In case you're too busy:
111
112
121
122
211
212
221
222

4. ## Re: coin tossed

1/2 laying it out on a tree diagram helped me.
its .375/.75 =.5

5. ## Re: coin tossed

Geesh Marko...keep it simple!

111 n/a
112 *
121 *
122
211 *
212
221
222 n/a

3/6

6. ## Re: coin tossed

Originally Posted by DenisB
Geesh Marko...keep it simple!

111 n/a
112 *
121 *
122
211 *
212
221
222 n/a

3/6
The original post specifies that he is required to use the rule $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$. (S)he was keeping it simple.

7. ## Re: coin tossed

Originally Posted by SlipEternal
The original post specifies that he is required to use the rule.
Is that the way you took it?
Actual wording shown: "i am using the conditional probability rule".
I took it as HIS decision.
A thousand apologies of which you may have one

8. ## Re: coin tossed

Originally Posted by SlipEternal
The original post specifies that he is required to use the rule $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$. (S)he was keeping it simple.
Here is a YouTube I have often used.
Note the form of conditional probability is more than $\dfrac{P(A\cap B)}{P(B)}$.

In this particular problem I bet the point is that the given is meant to be used to limit the space.