Question

A biased dice is such that, the probability of getting ‘k’ on the dice is k times the probability of getting 1. What is the probability of getting an odd number?

Answer 1

3/6

Answer 2

6 sided normal dice ?

Answer 3

idk

Answer 4

but in options there are answers 1/7,3/7..etc no answer is 6 in denominator

Answer 5

idk whether options are wrong

Answer 6

First, a dice has six sides, and the outcomes are mutually exclusive.

So, that means P[1] + P[2] + ... + P[6] = 1.

We are given that the probability of P[k] = k * P[1]

So, replace P[k] with k * P[1] in the first equation, and solve for P[1].

Then, you have the probabilities for the other outcomes. Then, solve P[1] + P[3] + P[5].

Answer 7

3/6?

Answer 8

k*p[1]+

Answer 9

I haven't did the math. I'm waiting for you to try to do it. I gave you the steps you would need to take to solve it.

Answer 10

k*p[1]+k*p[3]+k*p[5]/6

Answer 11

Ok, first, we need to find out what P[1] is. We don't know what that is yet, but we can figure it out.

We know that P[k] = k * P[1]. And, we know that P[1] + P[2] + ... + P[6] = 1. So, replace P[k], k != 2, with k * P[1].

P[1] + 2 * P[2] + 3 * P[3] + ... + 6 * P[6] = 1

Can you do the math for that?

Answer 12

Ooops, mistake.

Answer 13

I meant to type: P[1] + 2 * P[1] + 3 * P[1] + ... + 6 * P[1] = 1

Answer 14

That is just P[1] * (1 + 2 + 3 + 4 + 5 + 6) = 1, so P[1] is what?

Answer 15

1

Answer 16

No, let P[1] = p. Then, we have the equation:

p*(1+2+3+4+5+6)=1. Solve for p.

Answer 17

That step is just algebra. If you are being given these kind of problems, they would expect you to have some background in basic algebra.

Answer 18

First, sum the integers 1+2+3+4+5+6. What is that sum?

Answer 19

21

Answer 20

So, 21p = 1

Answer 21

21*p=1

Answer 22

p=1/21

Answer 23

Yes. So, P[1] = 1/21. What is P[2]?

Answer 24

2/21

Answer 25

9/21

Answer 26

Yes, so, to find out P[odd number], what's next?

Answer 27

3/7

Answer 28

Show me your work. How did you get 3/7?

Answer 29

1+3+5/21=9/21=3/7

Answer 30

Good.

Answer 31

oh i understood they given k times so we are using 2*p[2],3*p[3] etc right

Answer 32

No, P[3] = 3 * P[1].

Answer 33

As given in the problem. That is why we first had to find out what P[1] was, using the knowledge that the exhaustive sum of the mutually exclusive probabilities must sum to 1.

Answer 34

yeah.i made a mistake

Answer 35

Understood. Good, I think you understand the problem.

Answer 36

okay.can u tell me any reference for this kind of problem.means forum,books for probability

Answer 37

I think Googling something like "introduction probability" would be a good start. There are certainly many good books on this, also.

Answer 38

I have to leave for ~10 minutes, bbiab.

Answer 39

ok.bye i will post another question