Question

Thank you!

Answer 1

@mathslover

Answer 2

Thank you for helping @mathslover

Answer 3

I will try to help you at my level best.

Answer 4

ok thanks

Answer 5

Is 2/5 correct? or ?

Answer 6

You still there?

Answer 7

@mtbender74

Answer 8

Yeah...i'm looking at it. been a long time since i did probability.

Answer 9

Sorry to ask you again. I didn't want to bother you anymore :( @mtbender74

Answer 10

no bother...just dusting off the probability cobwebs

Answer 11

hahha ok. Thanks though.

Answer 12

Have another if this one is too difficult? @mtbender74

Answer 13

So here's where my mind is going...maybe if i type it out, it'll spark something.

The better way to answer a "at least" question is to turn it around. In other words, "At least 2 are late" is the same as "not exactly zero and not exactly one are late."

And those we can get at directly. All we have to do is find the probability that either none are late or exactly one is late and then subtract this probability from 1.

So, exactly zero late over 5 flights would be (7/10)^5. Let's just look at the numberator for now...7^5

For exactly 1 late (again only the numerator), we'd have (7^4)(3).

We add these together to get the exactly 0 *or* exactly 1 late...

7^5 + (7^4)(3) = 24010

The denominator would be 10^5 = 100000...

So we subtract 100000 - 24010 = 75990...still over 100000

The trouble is 75990/100000 = 7599/10000 is not an answer given.

Answer 14

WOW! I am going to read it just sec.

Answer 15

ok I see the problem. What do I do then? How do I solve?

Answer 16

Oh wait wait! Let me scan something and show you. It is my fault!@mtbender74

Answer 17

#11 and #12 are the ones I need help on that are marked open study.. But the chart table I didn't put it on here. Sorry. @mtbender74 see attachment

Answer 18

ok...that makes a *lot* more sense :)

So count the number of boxes that have at least 2 of 7, 8, or 9. There are 20 boxes total.

Answer 19

ok

Answer 20

11?

Answer 21

For #12, you have to remember to keep each die separate. typically, we imagine that they are colored red and green so that a R4 (a 4 on the red die) and a G5 (5 on the green die) is not the same outcome as R5 G4 even though they both show the same in real life.

Given that, there are 4 ways you can show 9 on a pair of dice (if you know the game craps, this is very important to know :)

R3 G6
R4 G5
R5 G4
R6 G3

There are 36 combinations possible for a pair of dice (this is your denominator)...

Answer 22

Yes...11/20 for the first one :)

Answer 23

Ok thanks and I printing out your notes so I can figure out others like it. Thank you very much for your help.

Answer 24

Very welcome... :) Just remember, probability is all about counting

Answer 25

I am reading #12 now.

Answer 26

ok thanks

Answer 27

Is it 1/18 then? For #12 ? @mtbender74

Answer 28

Oh wait is it 3/4 then??

Answer 29

Or am I way off :(

Answer 30

not quite..number of success on top and total on bottom

4 successful choices out of 36 total...4/36 = ?

Answer 31

1/9

Answer 32

Much better :)

Answer 33

ahha thanks. I promise no more questions for you today from me. Thank you You helped a lot and to understand it and not just give me the answers right away. @mtbender74