AH! I got a math test tomorrow and i need help! URGENT Computing Probabilty with combinations, Quality Control

Question

Alright i just need to know how to do questions such as

"If four letter code words are generated at random using the letters A,B,C,D,E, and F, What is the probability of forming a word without a vowel in it?"

And

"A box containing one dozen light bulbs includes 2 defective bulbs. A random sample of 4 bulbs is selected by the manufacturer. What is the probability that the sample contains

a. No defective bulbs?
b. One defective bulb?
c. Two defective bulbs?
D. At least one defective bulbs."

I really am looking for an explanation of how to get the answer, rather then just a simple answer but ANY help is appreciated!

The test is tomorrow so please reply ASAP.

Thanks!

Answer 1

1. In forming words, you are generally sampling WITH replacement, unless you are specifically told that repeated letters are not allowed (or in problems like "How many words can be formed from the letters of DAFFODILS..." where it is implied that you can only use each letter as often as it appears in the source word).
To form a word without a vowel, each letter must be one of the four consonants. So we have 4^4 acceptable combinations out of 6^4 possibilities, so the answer is 4^4 / 6^4 = (2/3)^4 = 16/81.

2. We're sampling without replacement here, and order isn't important, so we're looking at combinations. I'll write combinations in the form C(n, r) for n choose r.
We are choosing 4 items from 12 in all, so there are C(12, 4) = 12.11.10.9/4.3.2.1 = 495 combinations.
a) No defective bulbs: we must choose 4 bulbs from the 10 good ones, so we have C(10, 4) = 210 possible combinations, for a probability of 210/495 = 14/33.
b) One defective bulb: we must choose 3 bulbs from the 10 good ones and 1 bulb from the 2 bad ones, so we have C(10, 3) . C(2, 1) = 120 . 2 = 240 combinations, for a probability of 240/495 = 16/33.
c) Two defective bulbs: we'd expect this to be 1 - 14/33 - 16/33 = 3/33 = 1/11, but let's check: we need to choose 2 from the 10 good bulbs and 2 from the 2 bad ones, so we have C(10, 2).C(2, 2) = 45.1 = 45 combinations for a probability of 45/495 = 1/11.
d) You can get this either as the complement of (a), i.e. 1 - 14/33 = 19/33, or by adding (b) and (c) together, i.e. 16/33 + 1/11 = 19/33. In general the first method is better, because it requires less calculation, particularly when there are larger numbers involved (if there were 5 bad bulbs, we wouldn't want to calculate the probability of getting 1, 2, 3, 4 and 5 separately just to get this answer, when we only need to calculate the probability of getting 0 and subtract it from 1).

Answer 2

1. Your chance of obtaining four consonants in 4 picks (with replacement) is (2/3)^4 or 16/81. You don't worry about combination, as the only criterion involved is that a letter is not a vowel, and you can pick the same consonant more than once. If this latter condition is not so, the probability will change.
2. This is a "without replacement" situation, for once a bulb is picked out, it isn't available to be picked again. The number of different picks that can be made is 12!/(4!8!) or 495. The picks that none will be defective is
10!/(4!6!)= 210
The picks that one will be defective is
(2!/1!1!)(10!/3!7!)= 240
The picks that two will be defective is
(2!/2!0!)(10!/2!8!)= 45
Since 210 picks yield no defects, 285 picks must yield one or more defective bulbs. The probabilityies will then be a: 210/495=14/33
b/ 240/495= 16/33 c: 1/11 and d. 19/33

Answer 3

Hey! I love math... I'm a nerd, you don't need to tell me. Here are the notes I took on probablity.

If you are dealing with an "and" you multiply. If you roll two dice and you want a 1 AND a 2 you multiply 1/6 *1/6* 1/6 * 1/6.
If you are dealing with an or, you add. You role two dice and you want a 1 OR a 3 so you add 1/6 + 1/6.

In the first problem you have there I would start by finding the probablity of each letter bing in the four letter combination. Remember that the order doesn't matter. I hope this helps.

Answer 4

1. There are 6x5x4x3=360 possible four letter words. Note that order is important here. If we omit vowels, there are 4x3x2x1=24 possible words. So probability is 24/360.

2. This is a binomial distribution problem with n=4, p=2/12=1/6.

P(X=0)=(4C0)((1/6)^0)((5/6)^4)
P(X=1)=(4C1)((1/6)^1)((5/6)^3)
P(X=2)=(4C2)((1/6)^2)((5/6)^2)

P(at least 1)=1 - P(X=0)

Answer 5

1. Well there are 4 consonants
so 4nCr4/6nCr6 Thats assuming you cannot repeat letters
1/15

2. a. 10nCr4/12ncr4
b. 2nCr1/12nCr4
c. 2nCr2/12nCr4
d. in a we found none, so 1-10nCr4/12ncr4

Answer 6

1. use bcd and f
so its 4x4x4x4