Probability help

Hedgecutter

Wondering if my solution is right or even close ?

Hedgecutter

Also this one

spock.

1st question:

A. There are 6 ways a total of 7 can be obtained (16, 25, 34, 43, 52, 6,1) so 6/36 = 0.166
B. 2 ways 11 can be obtained (56, 65) = 2/36 =0.055
C. Add your 1st 2 answers
.1666 +. 055 = 0.222
D. You have the probability of it being 7 or 11, so the probability that it is neither is 1-(.222) = 0.777

spock.

The 2nd one is right except I think you misread the question. 50 girls have distinctions not 40

Hedgecutter

Thanks misread it alright, dice question is trick enough. Can't get my head around it.

spock.

It makes more sense if you have a table with all the possible outcomes

barneystinson

Hedgecutter wrote: »

Thanks misread it alright, dice question is trick enough. Can't get my head around it.

If you roll 2 dice there's only 36 possible outcomes.

11
12
13
14
15
16
21
22
23
24
25
26
31
32
33
34
35
36
41
42
43
44
45
46
51
52
53
54
55
56
61
62
63
64
65
66

Now, see how many add to 7?

And to 11?

That's your probability.

Hedgecutter

Thanks guys makes sense now. Cheers

Hedgecutter

Finding probability hard to get my head around. Looking at this question id say it's fairly simple but I'm struggling.

Yakuza

i) Six defectives doesn't appear as a possible outcome, and the sum of the other probabilities adds to one, so p(6 defectives) = 0.
ii) Looks good.
iii) You can do this two ways - P("one or more" defectives) is the same as 1 - P("Zero defectives"). Or you could just add up the independent probabilities P(1 defective) + P(2 defectives) + P(3 defectives) + P(4 defectives) + P(5 defectives).
iv) Two or fewer defectives = P(0 defectives) + P(1 defective) + P(2 defectives)
v) Like iii above, this can be done two ways - P(2 or more) = 1 - (P(0 defectives) + P(1 defective))
or P(2 or more defectives) = P(2 defectives) + P(3 defectives) + P(4 defectives) + P(5 defectives). You will end up with the same result.

You seem to be confusing independent events with mutually exclusive events. You only multiply probabilities when working out the probability of a combined, independent event for example:

What is the chance of getting heads when throwing a coin and getting a six when throwing a dice:
This is 1/2 * 1/6 or 1/12.

Getting back to your example, you are given that there are six particular possible outcomes. Each outcome is "mutually exclusive". This means that only one of the outcomes can happen (i.e. a sample can't have two and three defectives at the same time).

So, the probabilities of something happening sum to 1 (the batch has 0, 1, 2, 3,4 or 5 defectives).
In order to answer the later questions, you need to think about what must happen to satisfy the conditions in the question.

Take part iv) for example. In order for there to be two or fewer defectives, what must have happened? There must be either zero, one or two defectives. What are the chances of this happening? It's the total of the individual probabilities of these events taking place, or in this case (0.16 + 0.28 + 0.3 or 0.74 in total).
You can also answer it by taking the probability of what's called the "Complement" of the event and taking this away from 1 (since the probability of something happening is 1). Complement of the event is just fancy talk for "the opposite". In this case the opposite of the event "Two or fewer defectives" is "three or more" - which is (0.16 + 0.08 + 0.02 or 0.24 in total. Taking this from 1 yields 0.74 as before).

I hope I've explained it somewhat for you. Good luck!

Hedgecutter

Thanks. Once I put the kids to bed I'll have a read and another go. Cheers again.

Wacker The Attacker

My understanding of the first set of questions is as follows.

The possible number of combinations of the two dice spinning is 36 (6*6)
This remains constant and will form part of all four answers

(1) There are three ways of making 7 with two dice

6+1 (* 2 as we have 2 dice 6 and 1 and 1 and 6)
5+2 (* 2 as above)
4+3 (* 2 as above)

Therefore, your probability is 6/36 or 1/6

(2) There is only one way of making 11 with two dice
6+5 (* 2 as we have 2 dice 6 and 5 and 5 and 6)

Therefore, your probability is 2/36 or 1/18

(3) To determine the likelihood of either 7 or 11 coming you simply
add together the answers of the first two questions:

6\36 + 2\36 = 8\36 = 2\9

(4) To establish this we simply subtract 2\9 from 1 (as the answer can only ever add up to 1 or 100%)

Yakuza

I think it's just easier to state there are 6 ways to make a total of 7 and 2 ways to make a total of 11 with 2 dice;

D1 D2
 1 6
 2 5
 3 4
 4 3
 5 2
 6 1

D1 D2
 5 6
 6 5

Hedgecutter

Wondering if the question below is right?. Similar to the last one

FreaksAndGeeks

For part c) i believe the answer should be (0.34 x 0.34) because you are calculating the probability that the first sample has one defective AND that the second sample has one defective.

For part d) it is definitely wrong has having a probability greater than one is impossible.

So the probability of the first sample having two or less defectives is (0.25 + 0.34 + 0.23) but then you multiply this by the probability of the second sample containing not more than one defective which is (0.34 + 0 .23)

So the answer is (0.25 +0.34 + 0.23) x (0.34 x 0.23).

Yakuza

a and b are correct.
For parts c and d, you are looking at two independent events (1st event in c is one defective in first sample, 2nd event is one defective in second sample).

See my earlier post above on how to calculate probabilities of two independent events.

You are almost correct in that you have the calculated the correct probabilities of each of the individual events, but to work out the probability of both events happening together (fancy talk for this is the "joint event") you don't add the probabilities of the individual events, you....hint

multiply them!

Have another bash at c and d and post your answers.

Hedgecutter

I had another go.

spock.

Hedgecutter wrote: »

I had another go.

They're both right. Well done.