please help me with these questions on probability? not sure if my answers are right?

1) A jar contains 4 red discs, 6 white discs and 5 green discs. Three discs are removed at random from the jar, one after the other. Once a disc has been removed it is not replaced in the jar.

a) Find the probability that

(i) The first disc is red,

(ii) The second disc is white if the first disc is green.

b) Calculate the probability that

(i) Exactly two of the three discs are red,

(ii) The three discs are the same colour.

a) i ) 4/15

ii) 5/15 * 6/14 = 1/7

b) i) 4/15 * 3/14 = 2/35

ii)6/15 * 4/14 = 4/35

5/15 * 4/14 = 2/21

2/35 + 4/35 + 2/21 = 4/15

ii) red = 4/15 * 3/14 * 2/13 = 4/455

white = 6/15 * 5/14 * 4/13 = 4/91

green = 5/15 * 4/14 * 3/13 = =2/91

4/455 + 2/91 + 4/91 = 34/355

2) Two letters are posted on the same day. Letter A is sent by first-class post and has a probability of 0.9 of being delivered the next day. Letter B is sent by second-class post and has a probability of only 0.3 of being delivered by the next day.

a) Find the probability that

(i) both letters are delivered the next day

(ii) neither letter is delivered the next day

(iii) at least one of the letters is delivered the next day.

b) Given that at least one of the letters is delivered the next day, find the probability that letter A is delivered the next day.

i need to know if i've done these right, and if not then how do i get the correct answer? thanks

i have worked out:

a) i) 0.09 * 0.03 = 0.027

ii) 1 - 0.027 = 0.973

iii) 0.9 + 0.03 = 0.93

b) P( A | B ) = P (both) / P (A)

0.93 = 0.027 / P (A)

P(A) = 0.027/0.93 = 0.29..

3) Shahid, Tracy and Dwight are friends who all have birthdays during January. Assuming that each friend’s birthday is equally likely to be on any of the 31 days of January, find the probability that

a) Shahid’s birthday is on January 3rd

b) both Shahid’s and Tracy’s birthdays are on January 3rd

c) all three friends’ birthdays are on the same day

d) all three friends’ birthdays are on different days.

The answers i have worked out are as follows, but i'm not sure if they are right. could someone check, and if they aren't, could you please tell me how to do them? thank you

a) 1/3

b) 1/3 * 1/3 = 1/9

c) 1/3 * 1/3 * 1/3 = 1/27

d) 1 - 1/27 = 26/27

Thank you and sorry for the very long post :(

Re: please help me with these questions on probability? not sure if my answers are ri

1.

a) (i) Is correct

(ii) This asks "The second disc is white if the first disc is green" *not *"The second disc is white AND the first disc is green" you don't need to include the chance of getting a wite disc first because it states that you did in fact get it first. Just find the chance of getting a green disc when there are 14 discs

b) (i) Only 2 of the discs are red and 1 disc is not red. If R is red and O is other, the disks could be picked in the orders RRO, ROR, ORR. Find the probability of each happening and add them up.

(ii) Is correct

2.

In a lot of 2 you used 0.09 instead of 0.9 and 0.03 instead of 0.3

a) i) Your method is right

ii) You found the chance that both don't come the next day but that includes the chance of one coming and the other not.

iii) One or more letters arriving is equal to the chance of [neither arriving] Not happening. You know the chance of neither arriving from ii)

b) I'm not sure about this 1

3. a) Use 1/31 instead of 1/3

b) Same as a)

c) Find the chance that they are all on any given day, say... January 4th. There are 31 ways in which they can all be on the same day so multiply the chance of them all being on January 4th by 31.

d) You want to find the chance they are all on different days. There are 31 days which Shahid's birthday can be on; there are 30 days that Tracy's birthday can be on without it being on the same day as Shahid's. There are 29 days which Dwight's birthday can be on without it being on the same day as either of the others. Therefore there are 31*30*29 ways in which their birthdays can be different.

In total there are 31*31*31 ways you can choose their birthdays to be arranged.

Find the chance that their birthdays are different.