Question

Explain please !

Two dice are tossed, one after the other, in how many different ways can they fall ?

list the number of different ways the sum can equal

a) 3 b) 11

Answer 1

eh... there's more possibilities than 11...

Answer 2

think about it, 1,1 1,2 1,3 1,4 1,5 1,6 etc...

Answer 3

wait

Answer 4

well there is two question the first is a permutation, but i don't know how to aproach

Answer 5

Oh right, I thought that was a multiple choice...

Answer 6

I thought that too... So what is this asking then? Could you describe it more?

Answer 7

nick, he's asking how can the numbers add up to 3 and 11 in different cases.

Answer 8

oh, so 1-2, 2-1, 5-6, etc...?

Answer 9

yes that is the second one

Answer 10

My permutations suck.... I forgot atleast half of what I learned for permutation.

Answer 11

i think the first one is P = 12!/(12-2)! AM I RIGHT ?

Answer 12

sorry can't help you with that... I'm in trigonometry right now and I completely forgot about permutations I'm so sorry

Answer 13

so you are good in probability ?

Answer 14

well thanks anyway, nicklife2

Answer 15

probability- 2-1, 1-2, about thats it for a) if that's what you're asking?

Answer 16

Slot machine A standard slot machine contains three reels,
and each reel contains 20 symbols. If the first reel has five
bells, the middle reel four bells, and the last reel two bells,
find the probability of obtaining three bells in a row.

Answer 17

this one is of probability

Answer 18

someone help !

Answer 19

a) possiblities to get 3 are

1 and 2 or
2 and 1

Answer 20

is a question of permutation, how do you get there ? can you show me ?

Answer 21

are the different ways can they fall,expressed by : P = 12!/(12-2)! ?

Answer 22

i'm stuck :/

Answer 23

the formula you gave is the number of permutations of 2 in 12 which works out to 12*11 = 132

Answer 24

1. Two dice are tossed, one after the other, in how many different ways can they fall ? (first question)

Answer 25

2. list the number of different ways the sum can equal (second question) already solved

a) 3 b) 11

Answer 26

yes - i'm afraid probability is not my strong point - i'll have to think about it….

Answer 27

ok

Answer 28

Slot machine A standard slot machine contains three reels,
and each reel contains 20 symbols. If the first reel has five
bells, the middle reel four bells, and the last reel two bells,
find the probability of obtaining three bells in a row.( I have to solve 1. and 3.)

Answer 29

3. \[P(A) = (E _{_{1}})U(E _{2})U _{?}(E_{3})\]

Answer 30

the answer to the first one is 6*6 = 36
because your can get 1&1, 1&2, 1&3 1*3 etc - 6 ways
then 2&1, 2&2 etc

so that total of 36

Answer 31

E1 = 5/20 + 4/20 + 2/20

Answer 32

I MEAN P(A) = the expression above

Answer 33

althought I think is an intersection

Answer 34

i think those probabilities should be multiplied as each event \(result in each reel) is independent

so its 5/20 * 4/20 * 2/20

Answer 35

should the symbol be intersect not U? i.e. upside down U

Answer 36

That make sense, I'm no so good in prob

Answer 37

yea - if we added those numbers far too many people would win!!!

Answer 38

so the expression for the first one would be: \[P ^{r}\]

Answer 39

where P= is the number of possibilities and r= # of dice

Answer 40

maybe - i'm not too sure of all this notation in probability

Answer 41

1 a is 1.2 and 2,1
1 b is 5,6 and 6,5 if i understand the question correctly

Answer 42

P(A) = \[E _{1} *E_{_{2}} *E_{3}\]

Answer 43

yes

Answer 44

Hey man thanks a lot !

Answer 45

yw - (i'm a woman though , but thats ok)

Answer 46

i have a test of this but i think i struggle too much

Answer 47

oh I'm sorry
i didn't meet (until now) a woman so good in math

Answer 48

;) thanks again