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

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SUMMARY

This discussion focuses on calculating the probability of outcomes from tossing biased coins, specifically coin A with a probability of Heads at 0.6 and coin B at 0.25. The problem involves determining the probability of obtaining two Heads and one Tail from three tosses of a randomly chosen coin. The correct formula to use is P = (1/2) * 3 * (H_A)^2 * (T_A) + (1/2) * 3 * (H_B)^2 * (T_B), which incorporates the selection probability of each coin. The discussion emphasizes the importance of understanding conditional probabilities and suggests using a probability tree for verification.

PREREQUISITES
  • Understanding of binomial distribution
  • Familiarity with geometric distribution
  • Knowledge of conditional probability
  • Basic probability theory concepts
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  • Study the binomial distribution formula and its applications
  • Learn about conditional probability and Bayes' theorem
  • Explore probability trees and their use in complex probability problems
  • Investigate the differences in outcomes when replacing coins versus not replacing them
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Students and educators in statistics, mathematicians focusing on probability theory, and anyone interested in understanding the complexities of biased coin toss outcomes.

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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