How Do You Calculate the Probability of Coin Toss Outcomes with Biased Coins?

AI Thread Summary
The discussion focuses on calculating the probability of getting two Heads and one Tail from two biased coins, A and B, with probabilities of Heads at 0.6 and 0.25, respectively. The user seeks clarification on incorporating the probability of selecting either coin into their calculations. They derive a formula that combines the probabilities of each coin and the outcomes of the tosses. The conversation emphasizes the importance of understanding conditional probabilities and suggests checking the derived formula against the basic probability relationship. Overall, the user gains insights into constructing the problem without relying on visual aids.
AntSC
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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. Homework Statement

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.

Homework Equations

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?
 
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What is the probability you picked coin A?
 
A half
 
So if you picked a coin at random and tossed just once, what is the probability the result is a head?
 
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?
 
You can check it with a probability tree if you are unsure.
 
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 :)
 
AntSC said:
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="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##.
 
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