Two balanced coins are tossed. What are the expected value and the variance of the number of heads observed?
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I'm confused on how to start. I don't know how many times im suppose to toss coin, and how to set up my table for the expected value and then find variance, I know how to find expected value when I know how to use what's given to me, hope you can help. I feel that the answer is probably simple but i'm stuck on info i think it lacks.
p = 1/2 for heads or tails
this is a binomial distribution
mean = n p = n/2
variance = n p (1-p) = n/4
It is indeed lacking n so has to be answered that way
Wait a minute, they said two coins. Assume each tossed once. n = 2
thanks damon for your help
No problem! I can help you with that.
To find the expected value and variance of the number of heads observed when two balanced coins are tossed, we can use the concept of probability.
Let's start by understanding the possible outcomes of tossing two coins. When tossing one coin, there are two possible outcomes: heads (H) or tails (T). So, when tossing two coins, the possible outcomes are:
- HH (both coins land on heads)
- HT (first coin lands on heads, second coin lands on tails)
- TH (first coin lands on tails, second coin lands on heads)
- TT (both coins land on tails)
Now, let's find the expected value. The expected value is the average or mean value we would expect to observe over many repetitions of an experiment.
For each outcome, we can assign a value of 0 or 1, where 0 represents the outcome of tails and 1 represents the outcome of heads. So, the values for each outcome are:
- HH: 2 heads, value = 2
- HT: 1 head, value = 1
- TH: 1 head, value = 1
- TT: 0 heads, value = 0
To find the expected value, we multiply each outcome's value by its probability and sum them up. Since each coin is balanced, the probability of getting heads (H) or tails (T) is 1/2.
Expected value = (2 * 1/2) + (1 * 1/2) + (1 * 1/2) + (0 * 1/2)
Expected value = 1 + 1 + 0
Expected value = 2
So, the expected value of the number of heads observed when two balanced coins are tossed is 2.
Next, let's find the variance. The variance measures the spread or dispersion of the values around the expected value. For a random variable X, the variance is given by:
Variance(X) = [(X1 - E)^2 * P(X1)] + [(X2 - E)^2 * P(X2)] + ...
Here, X1, X2, ... represent the possible values of X, and P(X1), P(X2), ... represent their respective probabilities. E represents the expected value.
For our case, X can take values 0, 1, or 2 (number of heads observed) with respective probabilities of 1/4 (HH), 1/2 (HT and TH), and 1/4 (TT).
Variance = [(0 - 2)^2 * 1/4] + [(1 - 2)^2 * 1/2] + [(2 - 2)^2 * 1/4]
Variance = 4/4 + 1/2 + 0/4
Variance = 1 + 1/2
Variance = 3/2
So, the variance of the number of heads observed when two balanced coins are tossed is 3/2 or 1.5.
In summary:
Expected value = 2
Variance = 3/2 or 1.5
I hope this explanation helps! Let me know if you have any further questions.