need help with another probability problem?

Question

olving the rolling of a die. Assume that the die is weighted so that the probability of a 1 is 0.3, the probability of a 2 is 0.1, the probability of a 3 is 0.2, the probability of a 4 is 0.2, the probability of a 5 is 0.1, and the probability of a 6 is 0.1. You roll the die until the sum of all numbers which have appeared exceeds 3. A random variable X is defined to be the number of rolls.

How many different values are possible for the random variable X?

i got 4

PART 2:

Fill in the table below to complete the probability density function. Be certain to list the values of X in ascending order.

Value of XProbability
1
2
3
4

Answer 1

See my answer to your other question. If you can get that one you will get this one. Similar concepts.

Answer 2

Probability problems can be tricky, but in this case you can easily write out all the possible consequences

lets list all the possible outcomes (I got this from writing a tree chart on a piece of paper. The most number of rolls required is 4.
To solve part 2, I would use tally up the probabilites of 3 of the easiest groups and let 1- sum of other prob be the aswser for the most difficult group. I'm all for learning a short cut to it.

4
5
6
(3,1)=4
(3,2)=5
(3,3) =6
(3,4)=7
(3,5)=8
(3,6)=9
(2,1,1)=4
(2,1,2)=5
(2,1,3)=6
(2,1,4)=7
(2,1,5)=8
(2,1,6)=9
(2,2)=4
(2,3)=5
(2,4)=6
(2,5)=7
(2,6)=8
(1,1,1,1)=4
(1,1,2)=4
(1,1,3)=5
(1,1,4)=6
(1,1,5)=7
(1,1,6)=8
(1,2,1)=4
(1,2,2)=5
(1,2,3)=6
(1,2,4)=7
(1,2,5)=8
(1,2,6)=9
(1,3)=4
(1,4)=5
(1,5)=6
(1,6)=7

Answer 3

X = 1 = one roll
---> P(X = 1) = P('4' or '5' or '6') = .2+ .1+ .1 = 0.4
X = 2 = two rolls
---> P(X = 2) = 1-P( ['1' & '1'] or ['1' & '1'] or ['1' & '2'] or ['2' & '1']) = 1-(.3x.3+ .3x.3+ .3x.1+ .1x.3) = 0.76
X = 3 = three rolls
---> P(X = 3) = 1-P( ['1' & '1' & '1'] or ['1' & '1' & '1'] or ['1' & '1' & '1'] ) = 1-(3x .3x .3x .3) = 0.919
-- -- --
Number of different values for X = infinite (the dice can be rolled as many times as you want), starting from 1.

X = 1, P(X = 1) = 0.4
X = 2, P(X = 2) = 0.76
X = 3, P(X = 3) = 0.919
X > 3, P(X > 3) = 1

Answer 4

in case you %. a blue first the risk is two/8, and the risk of figuring out on a 2d blue is a million/7 So the risk of figuring out on 2 blues is two/8 x a million/7 = 2 in fifty six through fact there are the comparable quantity of black balls as blue ones the risk of figuring out on 2 black ones is likewise 2 in fifty six in case you %. a white ball first the risk is 4/8, and the risk of figuring out on a 2d white is 3/7. So the risk of figuring out on 2 whites is 4/8 x 3/7 = 12 in fifty six So the possibilities of figuring out on 2 balls the comparable is two/fifty six + 2/fifty six + 12/fifty six = sixteen/fifty six = 2/7