republican democrat independent
male 46 39 1
female 5 9 0

1. randomly select one senator, what is the probability of getting a republican, given that a male was selected?

2. randomly select one senator, what is the probability of getting a male, given that a republican was selected?

3. randomly select 1 senator, what is the probability of getting a female, given that an independent was selected?

4. randomly select 1 senator, what is the probability of getting a democrat or independent, given that a male was selected?

5. randomly select 2 senators, what is the probability of getting a democrat female and a republican male?

So what do YOU think ?

1. 46/86?

So what do YOU think ?

Is it correct?

1. 46/86?

Correct - there are 46 republicans out of 86 males. Of course you can simplify that fraction.

What answers did you come up with for the other questions?

2. 46/86?

2. 46/86?

Correct numerator. But the denominator should be equal to the total number of republicans - what number is that?

Correct numerator. But the denominator should be equal to the total number of republicans - what number is that?

I meant, 46/51.

3. 0/1?

Yes - answers for 2 and 3 are both correct. It seems you have the technique down nicely. Looking forward to seeing what you get for number 5.

Yes - answers for 2 and 3 are both correct. It seems you have the technique down nicely. Looking forward to seeing what you get for number 5.

4. 39+1 over 46+39+1
40/86?

And I don't get question 5. please help!

4. 39+1 over 46+39+1
40/86?

Good.

and i dont get question 5. please help!

To get a dem female (DF) and a rep male (RM) means you can either (a) select a dem female followed by a rep male, or (b) select a rep male followed by a dem female. The probabiliy of (a) is


P (DF\ followed \ by \ RM)= \frac 9 {100} \times \frac {46}{99}


Note that the numerator of each fraction is the number of senators of appropriate persuassion, ane the denominator is the total number of senators to pick from.

Using this same technique you can calculate the probability of (b) - what do you get? Finally add P(DF\ followed \ by \ RM) + P(RM \ followed \ by \ DF) to get the total probability.

9x46/100x99= 23/550 therefore, (23/550)+(23/550)= 46/550. We can reduce to 23/275. The final answer should be 23/275. Correct?

9x46/100x99= 23/550 therefore, (23/550)+(23/550)= 46/550. we can reduce to 23/275. the final answer should be 23/275. Correct?

Yes - correct!

Thank you very much.

Question: the starting 5 players for the NY Knicks team have agreed to make charity appearance. If you must send 3 player to United Way event and the other 2 to Heart Fund event, how many different ways can you make the assignments?

5! / (2)!(3)!= 10? Used the nCr formula.

5! / (2)!(3)!= 10? used the nCr formula.

Yep.

winning power ball lottery requires you to select 5 numbers between 1 and 53 inclusive and, in a separate drawing, you must also select the correct single number between 1 and 42 inclusive. Find the probability of winning the jackpot?

I don't know where to start.

The numerator is the number of ways that you can select the winning 5 numbers times the number of ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additional ball can be drawn from 42.

The numerator is the number fo ways that you can select the winning 5 numbers times the number if ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additiobnal ball can be drawn from 42.

Can you please show me... I really confused on this.

I'll give you a different example that illustrates the thought process. What's the probability of dealing a royal straight spade flush from a deck of 52 cards? A royal straight spade flush consists of the ace of spades, king of spades, queen of spades, jack of spades, and 10 of spades. It doesn't matter what order the five cards are dealt in. The number of ways those five cards can be dealt is 5! The total number of possible ways to deal five cards is 52 x 51 x 50 x 49 x 48 = 52P5 (permutations of 52 cards 5 at a time). Hence the odds of dealing a royal straight flush is 5!/(52P5) = 0.0000385%.

The numerator is the number of ways that you can select the winning 5 numbers times the number of ways you can select the correct extra number. The denominator is the total number of possible ways that 5 balls can be drawn from 53, times the number of ways that the one additional ball can be drawn from 42.

53C5 x 42C1

53!x42/(53-5)!(5!)

(53x52x51x50x49)(42)/5x4x3x2x1
= 120,526,770?