## Probability Involving 3 Coins

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maroonblazer
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### Probability Involving 3 Coins

I'm trying to learn how to solve probability problems using what I'm learning about Combinations, Permutations and the Counting Principle.

Suppose we toss three coins. What are the probabilities we get:

I start by looking for the number of outcomes in the sample space. Since each toss is an independent event I use the counting principle to determine that the sample space is 2*2*2 = 8 outcomes.

For part a) I need to find the number of combinations where zero heads are chosen. I'm struggling with how to set up the Combination.

Hints?

maggiemagnet
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### Re: Probability Involving 3 Coins

What is the probability of not getting "heads" on the first toss?

What is the probability of not getting "heads" on the second toss? (Remember that these are independent events!)

So what are the probabilities of not getting "heads" for each of the four tosses?

For independent events, do you multiply or add the probabilities?

maroonblazer
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### Re: Probability Involving 3 Coins

maggiemagnet wrote:What is the probability of not getting "heads" on the first toss?

What is the probability of not getting "heads" on the second toss? (Remember that these are independent events!)

So what are the probabilities of not getting "heads" for each of the four tosses?

For independent events, do you multiply or add the probabilities?

Thanks for your reply. I approached the problem in that same way at first (.5 * .5 * .5 = $\frac{1}{8}$) however I'm trying to understand how to use what I've learned about Combinations and Permutations to solve the problem, since for more complex problems it likely won't be as simple as just asking the "not" question.

maggiemagnet
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### Re: Probability Involving 3 Coins

Since you're needing to find the probability of something not happening (four times in a row), I'm not sure how you'd go about answering that question without taking that into account.

What other method or formula did they give you to use? Thanks!

maroonblazer
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### Re: Probability Involving 3 Coins

maggiemagnet wrote:Since you're needing to find the probability of something not happening (four times in a row), I'm not sure how you'd go about answering that question without taking that into account.

What other method or formula did they give you to use? Thanks!

They've defined "Theoretical Probability" as the ratio of the number of elements in an event to the number of equally likely elements in a sample space.

From the text:
"For example, the sample space for tossing two coins, {(H, H), (H, T), (T, H), (T, T)}, has four equally likely elements and the event of landing two heads, {(H, H)}, has one element. The theoretical probability that two heads will show when two coins are tossed is 1/4."

So...
The event of "0 heads" equals the event of "three tails". There's only one element in that event so the ratio is 1/8.

For the event of "1 head", I count three different elements:
${(H,T,T),(T,H,T),(T,T,H)}$

I'm struggling to determine how you'd calculate that same number of elements in that event using combinations or permutations.

Martingale
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### Re: Probability Involving 3 Coins

this experiment follows a binomial distribution

$P(X=x)= {3\choose x}\cdot \left(\frac{1}{2}\right)^3$

where x is the number or heads