Who can help me with my math homework?

Who can help me with my math homework?
I cannot seem to figure out the answers to some of these math problems. HELP!

1. Three cards are selected at random without replacement from a well-shuffled deck of 52 playing cards. Find the probability that three cards of the same suit are drawn. (Round your answer to four decimal places.)

2. A customer from Cavallaro’s Fruit Stand picks a sample of 3 oranges at random from a crate containing 65 oranges, of which 4 are rotten. What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)

3. Electronic baseball games manufactured by Tempco Electronics are shipped in lots of 36. Before shipping, a quality-control inspector randomly selects a sample of 8 from each lot for testing. If the sample contains any defective games, the entire lot is rejected. What is the probability that a lot containing exactly 2 defective games will still be shipped? (Round your answer to three decimal places.)

4. A student studying for a vocabulary test knows the meanings of 16 words from a list of 26 words. If the test contains 10 words from the study list, what is the probability that at least 8 of the words on the test are words that the student knows? (Round your answer to three decimal places.)

5. Jacobs & Johnson, an accounting firm, employs 14 accountants, of whom 6 are CPAs. If a delegation of 3 accountants is randomly selected from the firm to attend a conference, what is the probability that 3 CPAs will be selected? (Round your answer to three decimal places.)

thanks

Suggestion by sammich
Haha do you go to Texas A&M? My friend just asked me for help on these exact problems.

1. ( 13 nCr 2 * 4 ) / ( 52 nCr 2 ) = .235

2. 1 – ( ( 61 nCr 3 ) / ( 65 nCr 3 ) ) = .176

3. ( 34 nCr 8 ) / ( 36 nCr 8 ) = .6

4. Not sure on this one, but I know it has this in it: ( ( 16 nCr 10 ) / ( 26 nCr 10 ) )
[you just have to figure out how to account for the at least 8 words part]

5. ( 6 nCr 3 ) / ( 14 nCr 3 ) = .055

Suggestion by Pengy
1. (4C1)*(13C3)/(52C3)

2.P(x>0) = 1-P(x=0) where x is rotten. 1- (61C3) / (65C3)

3.probability of not drawing a defective game out of 8 = (34C8) / (36C8)

4.P(x=10) + P(x=9) +P(x=8) = (16C8) / (26C8) + (16C7)(10C1) / (26C8) + (16C6)(10C2) / (26C8)

5. P(x=3) = (6C3) / (14C3)

Add your own answer in the comments!

What’s the answer to these math problems?
Im just wondering.. thanks

Four cards are selected at random without replacement from a well-shuffled deck of 52 playing cards. Find the probability that four cards of the same suit are drawn. (Round your answer to four decimal places.)

A customer from Cavallaro’s Fruit Stand picks a sample of 3 oranges at random from a crate containing 60 oranges, of which 4 are rotten. What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)

Electronic baseball games manufactured by Tempco Electronics are shipped in lots of 20. Before shipping, a quality-control inspector randomly selects a sample of 7 from each lot for testing. If the sample contains any defective games, the entire lot is rejected. What is the probability that a lot containing exactly 3 defective games will still be shipped? (Round your answer to three decimal places.)

A student studying for a vocabulary test knows the meanings of 16 words from a list of 24 words. If the test contains 10 words from the study list, what is the probability that at least 8 of the words on the test are words that the student knows? (Round your answer to three decimal places.)

Jacobs & Johnson, an accounting firm, employs 15 accountants, of whom 9 are CPAs. If a delegation of 4 accountants is randomly selected from the firm to attend a conference, what is the probability that 4 CPAs will be selected? (Round your answer to three decimal places.)

And how did you get them?

Thanks!!

Suggestion by trueprober
Hi Hi Hi, let me answer at least one of these. That is the allowed quota.
First one.
Four cards can be drawn out of 52 in 52C4 ways. Hence sample space would have 52C4.
Now out of 13 of same suit 4 can be drawn in 13C4 ways. This thing may happen to any one of the four suits. So the total in the event will be 4* 13C4.
HEnce the probability of getting four cards of same suit = 4*13C4 / 52C4.= 44/4165 = 0.0106.

Know better? Leave your own answer in the comments!