S1 Probability Coin Toss

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1. Feb 25, 2015

AntSC

Having trouble with certain binomial and geometric distribution questions, which is indicating that my understanding isn't completely there yet. Any help would be greatly appreciated.

1. The problem statement, all variables and given/known data

A bag contains two biased coins: coin A shows Heads with a probability of 0.6, and coin B shows Heads with a probability 0.25. A coin is chosen at random from the bag and tossed three times.
Find the probability that the three tosses of the coin show two Heads and one Tail in any order.

2. Relevant equations

3. The attempt at a solution
Probabilities:
$H_{A}=0.6$ and $T_{A}=0.4$
$H_{B}=0.25$ and $T_{B}=0.75$

Possibilities for 2 heads and one tail in any order:
$$3\left ( H \right )^{2}\left ( T \right )$$

Is this correct so far?
My question is how to incorporate the probability of picking coin A or coin B into the problem?

2. Feb 25, 2015

Simon Bridge

What is the probability you picked coin A?

3. Feb 25, 2015

AntSC

A half

4. Feb 25, 2015

Simon Bridge

So if you picked a coin at random and tossed just once, what is the probability the result is a head?

5. Feb 25, 2015

AntSC

Ah i see it now.
$$P=\frac{1}{2}3\left ( H_{A} \right )^{2}\left ( T_{A} \right )+\frac{1}{2}3\left ( H_{B} \right )^{2}\left ( T_{B} \right )$$
Is this right?

6. Feb 25, 2015

Simon Bridge

You can check it with a probability tree if you are unsure.

7. Feb 25, 2015

AntSC

Sure. I want to start to dispense with the need for visual aids and make sure i can construct the problem without.
Especially when dealing with a larger set of choices, like 52 cards. A tree then won't be so helpful.
Thanks for the dialogue. I think i needed to get it out there to help work it through.
You might see a few more questions from me in future :)

8. Feb 25, 2015

Ray Vickson

QUOTE="AntSC, post: 5021860, member: 450435"]Ah i see it now.
$$P=\frac{1}{2}3\left ( H_{A} \right )^{2}\left ( T_{A} \right )+\frac{1}{2}3\left ( H_{B} \right )^{2}\left ( T_{B} \right )$$
Is this right?[/QUOTE]

If $E$ is the event "2H, 1T (any order)", does your formula satisfy the basic relationship
$$P(E) = P(E|A) P(A) + P(E|B) P(B) ?$$
If it does, it is OK.

BTW: you might compare this with the scenario where you replace the coin after each toss and then ask about $E$.