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#1 2007-06-30 06:33:40

vcantrell01
Guest

URGENT Probability Problem!!!!

Please, I really need help with these:

1.
The numbers 1 to 10 are written on sheets of paper and placed in a hat.   
If two sheets are selected at random then find the following:
1)The probability both numbers are ODD with replacement.
2) The probability of selecting a 2 and then a 9, with replacement. 

2.
One card is selected from a deck of standard poker cards (assume a typical deck of 52 totals cards numbered from 2 thru 10 and Jack thru Ace) at random.  Find the probability the card is a spade OR less than 8 (that is, from 2 to 7 only).

3.
The State Pick-3 lottery uses 27 balls numbered 1 to 9 (3 each).   
What is the probability that my ticket numbered “222” will win ?   
Assume that the balls are NOT replaced after each selection.

Thank you verryy much in advance, hope I'll get some help.
It's due tomorrow

#2 2007-06-30 07:18:08

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: URGENT Probability Problem!!!!

1 is straightforward; you should be able to do it yourself. The answers are (i) 1⁄4 and (ii) 1⁄100.

2
The deck of cards has 13 spades and 24 cards less than 8 (namely cards 2–7 in each of the four suits). Of these, 6 are both spades and less than 8. Therefore the number of cards that are either spades or less than 8 (or both) is 13 + 24 − 6 = 31. So the probability of picking such a card is …?

3
If the balls are not replaced, the total number of ways to picking 3 numbers from the 27 balls is [sup]27[/sup]C[sub]3[/sub]. There is only one way to pick 222 as the winning number. So the probability that your ticket will win is …?

Last edited by JaneFairfax (2007-06-30 07:21:59)

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#3 2007-06-30 14:29:50

vcantrell
Member
Registered: 2007-06-30
Posts: 1

Re: URGENT Probability Problem!!!!

Thank you very much for your answers.

I didn't understand, so what would be the answer for number 3?
Thank you again.

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#4 2007-06-30 19:42:05

Laterally Speaking
Real Member
Registered: 2007-05-21
Posts: 356

Re: URGENT Probability Problem!!!!

3-- If you have only one chance of getting it right, that part's pretty simple. As for the total number of combinations, you could figure it out like this: for the first pick, there are 27 balls to chose from. For the second, there is one less ball to pick. For the third, there is yet another ball missing.


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