Probability (1 Viewer)

[jexca]

I've been working on this question:

Two dice are rolled and two coins are tossed. Find the probability of getting:
a) 2 heads and a double 6
b) a head, a tail and a total of 8
c) 2 heads and a double

wasn't sure how to approach it, i've never been too good with probability, but i drew up a table with 1-6 across the top and down the side, then did the possible results from each throw e.g.:
1 2 3 4 5 6
1 1,1 1,2
2 2,1
3 3, 1
4 4,1
5
6

etc etc.
then did a tree diagram for the heads and tails (i think i'm a visual learner lol).

Anyway, back to the point.

for part a), the probability of getting 2 heads is 1/4
the probability of getting a double 6 = 1/36

I've worked out the way they've got the answer in the book is by multiplying
1/4 x 1/36 = 1/144
though I always thought you multiple along the branches (of the factor tree) then add the results.

Same deal with the other questions, multiplying when I thought you had to add.

Can anyone explain to me why it happens like this? :S i'm really confused now as to which questions I add and which ones I multiply.

Thanks so much, and all the best for monday.

 

[watatank]

P (A and B) = P(A) * P(B)

P(A or B) = P(A) + P(B)

take the first question for example. getting 2 heads and a double 6 are two independent events so you multiply the probabilities of both happening. just like you did to calculate the probability of getting two heads (1/2 * 1/2) and two sixes (1/6 * 1/6) separately. the probability of one event or the other you add them.

keep referring to the same graph and the table you made to do the other questions. the other answers are

b) 1/2 * 5/36 = 5/72

c) 1/4 * 1/6 = 1/24

what table did you make? it should look something like this

....| 1 | 2 | 3 | 4 | 5 | 6 |
1..|2 | 3 | 4 | 5 | 6 | 7 |
2..| 3 | 4 | 5 | 6 | 7 | 8 |
3..| 4 | 5 | 6 | 7 | 8 | 9 |
.
.
6 | 7 | 8 | 9 | 10 | 11 | 12 |

good luck for your exam