probability problem!?

Question

a die is loaded so that probability of face 'i' is proportional to i, i=1,2,............,6.what is the probability of an even no. occuring when the die is rolled?
& no, the answer is not 0.5

Answer 1

assuming propationality:

Probability(i)=P(i)= const * i= c*i

Total= c (1+2+3+4+5+6)=21 c=1

so, the constant : c=0.04762

then, for even numbers: 2,4,6

the probability= 0.04762 *(2+4+6)=0.5714

thanks
:)

Answer 2

1

Answer 3

I'm guessing that by 'proportional' you mean 'linearly proportional'.

The sum of the weightings for each face have to add up to 1. Just adding up the numbers from 1 through 6 we get 21 so the weighting 'functions' would be 1/21, 2/21, 3/21, etc and the sum of the even weights is
2+4+6 = 12 so the 'odds' of getting an even number should be 12 in 21. Which is considerably more than 1 in 2.


Doug

Answer 4

Add up all the number 1-6. The sum is 21. Then add up all the even numbers. The sum is 12. So the probability of getting an even number is 12/21 (Or 4/7 in its simplest terms, or 57.1%)
Homework tip: 4/7 is the answer your teacher probably wants.

Answer 5

The probabilities of all the possibilities of the die have to add to one.


1 = (1k) + (2k) + (3k) + (4k) + (5k) + (6k)
21k = 1
k = 1/21
P(roll a one) = 1/21
P(roll a two) = 2/21
P(3) = 3/21 = 1/7
P(4) = 4/21
P(5) = 5/21
P(6) = 6/21 = 2/7

P(even number) = P(2) + P(4) + P(6) = (2/21) + (4/21) + (6/21)
=12/21
=4/7

Answer 6

P(1)=1/(1+2+3+4+5+6) = 1/21
P(2)=2/21

P(x)=x/21

P(even)=P(2)+P(4)+P(6) = (2+4+6)/21 = 12/21 = 4/7

Answer 7

2+4+6=12
1+2+...+6 = 21

12/21=0.571

Answer 8

1*1+2*2+3*3+4*4+5*5+6*6=91
56/91=0.615

Answer 9

0.5