Question

The face of a 12 sided die are numbered from 1 to 12. What is the probability of rolling 9 AT LEAST TWICE in ten tries. Medal for answer

Answer 1

Can you find out how many ways the thrower will get 9 twice or more than twice times?

Answer 2

Hint : the user can get 9 exactly twice in 12 throws by 12C2

Answer 3

one sec

Answer 4

no, it only says 9 at least twice in 10 tries

Answer 5

doesnt say how many ways it can b thrown

Answer 6

I think it could be favorable outcomes / number of outcomes

Answer 7

its kind of like that, its a success/ failure probability

Answer 8

Plug in your #'s?

Answer 9

its more complicated than that

Answer 10

what is the probability of getting 0 nines

Answer 11

Do you have options to chose from?

Answer 12

no, but u can use binomial distribution to help you

Answer 13

P(at least 2) = 1- [P(0) + P(1)]
P is probability

Answer 14

is P(0)= 11/12?

Answer 15

i thought of that method too, but i didnt what probability of 0 nine is

Answer 16

you can find P(0) and P(1) with binomial distribution
(nCk)*(p^k)*(q^(n-k))
p is probability of succses = 1/12
q is probability of failure = 9/12

Answer 17

oops sory q = 11/12

Answer 18

Same thing as Success/Failure 9/12 and 11/12

Answer 19

I got zero for the answer, but its wrong

Answer 20

Did you get a decimal?

Answer 21

\[p(getting 9 inone throw)=\frac{ 1 }{ 12 }\]
\[P \prime=1-\frac{ 1 }{ 12 }=\frac{ 11 }{12 }\]
\[p(no 9 )=\left( \frac{ 11 }{ 12} \right)^{10}\]\[P \left( one 9 only \right)=\frac{ 1 }{12 }\left( \frac{ 11 }{ 12 } \right)^{9}\]

Answer 22

Required probability=1-[P(no 9)+P(one 9 only)]

Answer 23

Im not getting the right answer, what are you getting?

Answer 24

the answer should b 0.2003

Answer 25

what's your answer?
P(0) = (10C0)*((1/12)^0)*(11/12)^10
P(1) = (10C1)*((1/12)^1)*(11/12)^9

Answer 26

there we go :DD

Answer 27

but i dont get how you did that

Answer 28

can you explain it plz

Answer 29

C is combination
and for calculate (11/12)^9 and (11/12)^10 i use calculator :D

Answer 30

why did u put the exponent of 9 and 10

Answer 31

For those who own a TI-83 or TI-84 calculator and are familiar with probability distributions: type in the following: binomcdf(10,1/2,1). Result: 0.7997; subtracting this from 1 results in 0.2003, which is, as you say, the correct answer.

Answer 32

For those who don't own one of these calculators, or who must "show work," use the formula for binomial probability:

Answer 33

n is number of trials and k is number of possible ways to choose k "successes" from n trials
so n = 10 and k=0 for P(0) , k=1 for P(1)

Answer 34

P(k) = (nCk)*(p^k)*(q^(n-k))

Answer 35

hanifahrika is right on target, calculating, as she did, the probabilities of getting zero 9's in 10 trials and one 9 in 10 trials. Add these two probabilities together to get the probability of getting either zero or one 9's. To get the probability of getting at least 2 9's, subtract this sum from 1: 1-0.7997=0.2003.

Answer 36

is that the same formula as P(x)= nCr P^r q^n-r?

Answer 37

Yes, it is. I'd suggest typing p(x) = nCr * (p^r) * (q^[n-r]) for the sake of clarity.

Answer 38

Okay ty guys I understand it now

Answer 39

That's essentially the same as what hanifahrika provided. Delighted!