Prob Dist.. new question....

need help with b.

got 7.5 for mean and 2.37 for standard deviation.. where i go from here?

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- April 26th 2006, 09:54 PMArbiturProbibility Distribution
Prob Dist.. new question....

need help with b.

got 7.5 for mean and 2.37 for standard deviation.. where i go from here? - April 29th 2006, 11:23 AMRanger SVO
A card is drawn from a standard deck and replaced, if this experiment is repeated 30 times, what is the probability that

**no more**than 5 cards are spades.

There are 52 cards in a deck and 13 are spades. Our probability of drawing a spade on each draw is 1/4 and our probability of**not**drawing a spade is 3/4 but this is just 1 draw, we have 30.

Also the question says "no more than 5", so we need the probability for no spades, 1 spade, 2 spades, 3 spades, 4 spades and 5 spades.

No Spades --> We have 30 trials and want no spades, out of 30 draws this will happen only 1 way. Probability of getting a spade is (1/4)^0 ( raised to 0 because we want no spades) times the probability of no spades is (3/4)^30 (raised to 30 besause we want everthing else 30 times)

So (30 Cr 0)*(1/4)^0 *(3/4)^30 =**the probability of no spades**.

One Spade --> Out of 30 trails we can get 1 spade 30 different ways. The probability of getting a spade is now (1/4)^1 (raised to 1 because we want one spade) times the probability of no spades (3/4)^29 (raised to 29 because we need everything other than spades for the other 29 draws)

So ( 30 Cr 1 )* (1/4)^1 *(3/4)^29 =**the probability of 1 spade**

Get the Idea

Two Spades --> Out of 30 trials we can get 2 spades 435 different ways.

So (30 Cr 2)*(1/4)^2 * (3/4)^28 =**the probability of 2 spades.**

Three Spades

(30 Cr 3) *(1/4)^3 *(3/4)^27 =**the probability of 3 spades**

Four Spades

(30 Cr 4) *(1/4)^4 *(3/4)^26 =**the probability of 4 spades**

And finally Five Spades

(30 Cr 5) *(1/4)^5 *(3/4)^25 =**the probability of 5 spades**

Now we add the probabilities together to get the answer.

If anyone knows a simpler way let me know.