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cameron

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What is the probability that you get at least one red card?

You draw five cards at random from a deck of 12, 8 red and 4 white,replacing the drawn card before the next trial
**Probabilities must be in decimal form and rounded to 4 decimal places**

Posted 788 day ago

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Answers (3)

montie 		Let X be the number of red cards drawn. Because we are drawing with replacement X has the binomial distribution with n = 5 trials and success probability p = 8/12 = 2/3

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 5 , p = 0.6666667 )

the mean of the binomial distribution is n * p = 3.333333
the variance of the binomial distribution is n * p * (1 - p) = 1.111111
the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.054093

The Probability Mass Function, PMF,
f(X) = P(X = x) is:

P( X = 0 ) = 0.004115226
P( X = 1 ) = 0.04115226
P( X = 2 ) = 0.1646091
P( X = 3 ) = 0.3292181
P( X = 4 ) = 0.3292181
P( X = 5 ) = 0.1316872

P( X ≥ 1 ) = 1 - P(X = 0) = 0.9958848 ~= 0.9959

Posted 788 day ago

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Davido 		Easier to calculate the probability that all draws are white. Probability on each draw:

0.3333 for each. So (0.3333)^5

= 0.004115 (or 0.4%)

Answer for all red = 99.6%

Added in edit - to four places - 99.59%

Posted 788 day ago

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twocox 		The opposite of 'getting at least one red card' is 'getting 5 white cards'.
So P=1-(4/12)^5=1-(1/3)^5
P=1-1/243=0.9959

Posted 788 day ago

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