Six numbers writtien on separate cards.

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qcc

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You have the six numbers 1, 2, 3, 4, 5, and 6 written on separate cards in a hat. You choose cards one at a time, without replacing them, and write down the numbers in the order you choose them.
a) How many outcomes are there for this experiment?





b) What is the probability that your number is 54322?







c) Suppose you do this experiment 3000 times. Write down an expression for the probability that at least once you get the number 54322.







10. You choose four cards from the 26 hearts and diamonds in an ordinary deck. Order does NOT matter. What is the probability that the ace of hearts is among your four cards?
 
qcc said:
You have the six numbers 1, 2, 3, 4, 5, and 6 written on separate cards in a hat. You choose cards one at a time, without replacing them, and write down the numbers in the order you choose them.
a) How many outcomes are there for this experiment?

b) What is the probability that your number is 54322?

c) Suppose you do this experiment 3000 times. Write down an expression for the probability that at least once you get the number 54322.

10. You choose four cards from the 26 hearts and diamonds in an ordinary deck. Order does NOT matter. What is the probability that the ace of hearts is among your four cards?
Click to expand...
We need to see your work! Otherwise we don't know where you are getting stuck.

Are you SURE it says "without replacement"? If that is the case, then parts b) and c) should be obvious.

Often it is useful to replace "at least one" with "not zero," using the property that
P(none) = 1 - P(at least one.)

Show us how far you can get . . .
 
qcc said:
I got 6x4x3x2x1/5!= 6 outcomes. that is all i have.
Click to expand...
Why in the world divide by 5!? There are 6 ways to choose the first card, 5 ways to choose the second card and so on. Where does the divisor of 5! come from?

As Subhotosh Khan has implied, please check that you have written down the question exactly. The question as asked is trivially easy.
 
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