Probability help (1 Viewer)

[juampabonilla]

14. The letters of ENTERTAINMENT are arranged in a row. Find the probability that:
(b) two Es are together and one is apart,
(c) all the letters E are apart,
(d) the word starts and ends with E.
15. A tank contains 20 tagged fish and 80 untagged fish. On each day, four fish are selected
at random, and after noting whether they are tagged or untagged, they are returned to
the tank.
(d) What is the probability of selecting no tagged fish on exactly three of the seven days
during the week?
23. Four adults are standing in a room that has five exits. Each adult is equally likely to leave
the room through any one of the five exits.
d) What is the probability that no more than two adults come out any one exit?
24. (a) Five diners in a restaurant choose randomly from a menu featuring five main courses.
Find the probability that exactly one of the main courses is not chosen by any of the
diners.

 

[juampabonilla]

Any help is appreciated

 

[braintic]

14(b)
The non-Es are irrelevant (even though there are doubles).
No of ways of choosing positions for the Es is 13C3.
Block 2 Es. There are 12 positions for placing this block and the rest of the letters. No. of ways of placing the 3 Es is 12P2.
But there are 11 ways of getting the 3 Es in a block, and each of these 11 ways can be brought about in 2 ways using my method: EE/E or E/EE.
So probability is (12P2 - 22) / 13C3 = 5/13

(c)
Add back the 11 ways of getting 3 Es in a block, then take away from 1:
1 - (12P2 - 11) / 13C3 = 15/26

(d)
There are 11 ways of placing the Es so that 2 of them start and finish the word (the 11 positions for the 3rd E).
So prob is 11/13C3 = 1/26

15(d)
This is binomial probability (so not sure why it is in this Cambridge exercise). Using the answers for part (a) & (b):
7C3 (0.403)^3 (0.597)^4 = 0.291

23(d)
Using the answers to parts (a) & (c):
1 - 1/125 - 16/125 = 108/125

24(a)
(a copy and edit of my post a few weeks ago)

Assign diners to meals (not meals to diners).

The only way this scenario can occur is if 3 meals are chosen by exactly 1 person, 1 meal is chosen by 2 people, 1 meal is chosen by no-one.

Select the meals which are to be chosen by 0 or 2 people ... 5P2 ... order matters.

Select the two diners who are to assigned to the meal which you have picked for two diners ... 5C2

Arrange the other 3 diners amongst the other 3 meals ... 3!

Total number of ways the meals can be chosen ... 5^5

So 5P2 times 5C2 times 3! / (5^5) = 48/125

 

[laters]

24(a)
Number of ways for us to decide which meal is NOT chosen = 5
This leaves us with 4 meals for the diners to choose from. All the meals have to be chosen. But ONE meal has to be chosen exactly twice. Out of the 4 remaining meals, this can be chosen in 4 ways.
Let’s arrange the meals in order (i.e. 1st goes to diner 1, 2nd goes to diner 2 etc.) if the meals are A A B C D, this can be done in ways
Sample space =
P=

 

[juampabonilla]

Thank you guys, nice explanations.