Probability

[Guest]

I know I just registered and everything, but I have a math assignment and I'm stuck.

A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chosen. What is the probability that the value of the coins is at least 50 cents?

I've got all this confusing stuff writtien down. I started by saying (2C1*4C0*6C5)+(other combinations)
12C6

But from there I'm stuck, and I"m not even sure that's right.

Help please? :?

 

[Guest]

I've pretty much given up on that other approach. Too confusing for me.

This is what I wrote instead.

P|N|D
0|0|6
0|1|5
1|0|5
0|2|4

Those are the only 4 combinations that add to at least 50 cents.

Total combinations = 12C6 = 924

Therefore the probability of the coins adding to at least 50 cents is 4/924 which reduces to 1/231.

I'm hoping that's correct.

While I'm at it, the second part of the assignment is this:

"A letter is picked at random from the alphabet. Find the probability that the letter is contained in the word "house" or in the word "phone"."

My answer was 5/26 is the probability of a letter from the first word, plus the probability of a different letter from the second word which would be 2/26, bringing a total of 7/26. Did I do it correctly?

Thanks

 

[Denis]

your breakdown is correct; but the combinations total 127:

P|N|D
0|0|6 : 6C6 = 1
0|1|5 : 6C5 * 4C1 = 6 * 4 = 24
1|0|5 : 6C5 * 2C1 = 6 * 2 = 12
0|2|4 : 6C4 * 4C2 = 15 * 6 = 90

"Those are the only 4 combinations that add to at least 50 cents."

That's 4 arrangements, not combinations...

"Total combinations = 12C6 = 924" : correct!

"Therefore the probability of the coins adding to at least 50 cents is 4/924 which reduces to 1/231."

That's 127/924 (very close to 1/9)

While I'm at it, the second part of the assignment is this:
"A letter is picked at random from the alphabet. Find the probability that the letter is contained in the word "house" or in the word "phone"."
My answer was 5/26 is the probability of a letter from the first word, plus the probability of a different letter from the second word which would be 2/26, bringing a total of 7/26. Did I do it correctly?

Correct! house and phone use 7 different letters: h o u s e p n

 

[Guest]

your breakdown is correct; but the combinations total 127:

P|N|D
0|0|6 : 6C6 = 1
0|1|5 : 6C5 * 4C1 = 6 * 4 = 24
1|0|5 : 6C5 * 2C1 = 6 * 2 = 12
0|2|4 : 6C4 * 4C2 = 15 * 6 = 90

"Those are the only 4 combinations that add to at least 50 cents."

That's 4 arrangements, not combinations...

"Total combinations = 12C6 = 924" : correct!

"Therefore the probability of the coins adding to at least 50 cents is 4/924 which reduces to 1/231."

That's 127/924 (very close to 1/9)

While I'm at it, the second part of the assignment is this:
"A letter is picked at random from the alphabet. Find the probability that the letter is contained in the word "house" or in the word "phone"."
My answer was 5/26 is the probability of a letter from the first word, plus the probability of a different letter from the second word which would be 2/26, bringing a total of 7/26. Did I do it correctly?

Correct! house and phone use 7 different letters: h o u s e p n

Ah right, I get it now. I knew four seemed too small. Thanks a lot.