# help/confirmation on multi-stage question

• Oct 15th 2008, 01:55 PM
chrisc
help/confirmation on multi-stage question
consider a random experiment done in 2 stages
first 4 dice are rolled, then a coin is flipped as many times as the number 6 appears

a)find the probability of getting fewer than 2 heads
b)knowing the experiment resulted in fewer than 2 heads, what is the conditional probability that exactly 3 sixes were obtained in the first stage

here is what i have done

a)
probability of getting 1 head = (1/12)(11/12)(11/12)(11/12)
probability of getting 0 head = (11/12)^4
probability of getting fewer than 2 heads = 0.7703

b)
for this, i calculated the probability of getting a certain number of sixes (1, 2, 3 and 4) and then i multiplied those with the probability of getting fewer than 2 heads (respectively)

i finished with this

(5/1296)(1/216)/(0.01663) = 0.001074

I divided by the probability of each set of sixes that would give fewer than 2 heads, that is what the 4 significant digit number is.

I am somewhat confident in my answer to (a), but and Im not too sure about (b). Help, confirmation?
• Oct 15th 2008, 05:49 PM
mr fantastic
Quote:

Originally Posted by chrisc
consider a random experiment done in 2 stages
first 4 dice are rolled, then a coin is flipped as many times as the number 6 appears

a)find the probability of getting fewer than 2 heads
b)knowing the experiment resulted in fewer than 2 heads, what is the conditional probability that exactly 3 sixes were obtained in the first stage

here is what i have done

a)
probability of getting 1 head = (1/12)(11/12)(11/12)(11/12)
probability of getting 0 head = (11/12)^4
probability of getting fewer than 2 heads = 0.7703

b)
for this, i calculated the probability of getting a certain number of sixes (1, 2, 3 and 4) and then i multiplied those with the probability of getting fewer than 2 heads (respectively)

i finished with this

(5/1296)(1/216)/(0.01663) = 0.001074

I divided by the probability of each set of sixes that would give fewer than 2 heads, that is what the 4 significant digit number is.

I am somewhat confident in my answer to (a), but and Im not too sure about (b). Help, confirmation?

Let X be the random variable number of sixes.

X ~ Binomial(n = 4, p = 1/6).

Let Y be the random variable number of heads.

Y ~ Binomial(n = x, p = 1/2) where x is the possible values of X, that is, 0, 1, 2, 3, 4.

(a) $\displaystyle \Pr(Y < 2) = \Pr(Y = 1) + \Pr(Y = 0)$.

$\displaystyle \Pr(Y = 1) = \Pr(Y = 1 | X = 1) \cdot \Pr(X = 1)$ $\displaystyle + \Pr(Y = 1 | X = 2) \cdot \Pr(X = 2) + \Pr(Y = 1 | X = 3) \cdot \Pr(X = 3)$ $\displaystyle + \Pr(Y = 1 | X = 4) \cdot \Pr(X = 1) = \, ....$

$\displaystyle \Pr(Y = 0) = \Pr(Y = 0 | X = 0) \cdot \Pr(X = 0)$ $\displaystyle + \Pr(Y = 0 | X = 1) \cdot \Pr(X = 1) + \Pr(Y = 0 | X = 2) \cdot \Pr(X = 2)$ $\displaystyle + \Pr(Y = 0 | X = 3) \cdot \Pr(X = 3) + \Pr(Y = 0 | X = 4) \cdot \Pr(X = 1) = \, ....$

Therefore $\displaystyle \Pr(Y < 2) = \Pr(Y = 1) + \Pr(Y = 0) = \, ....$

The calculations are left for you to do.

----------------------------------------------------------------------------------------------------------------------

(b) $\displaystyle \Pr( X = 3 | Y < 2) = \frac{\Pr(Y = 1 | X = 3) \cdot \Pr(X = 3) + \Pr(Y = 0 | X = 3) \cdot \Pr(X = 3)}{\Pr(Y < 2)}$.

Substitute the required probability values from the calculations in part (a).
• Oct 15th 2008, 07:25 PM
chrisc
let me know if i am screwing up the conditional probability because i think i may be

a)

y=1:

(0.06419)(0.09645 + 0.01929 + 0.003858 + 0.09645)
=0.01387

y=0:

(0.7061)(0.4823 + 0.09645 + 0.01929 + 0.003858 + 0.09645)
=0.4931

Pr(Y<2) = 0.5070

--------

b)

(0.003858)(0.06419 + 0.7061)/(0.5070) = 0.005862
• Oct 15th 2008, 07:35 PM
mr fantastic
Quote:

Originally Posted by chrisc
let me know if i am screwing up the conditional probability because i think i may be

a)

y=1:

(0.06419)(0.09645 + 0.01929 + 0.003858 + 0.09645)
=0.01387

y=0:

(0.7061)(0.4823 + 0.09645 + 0.01929 + 0.003858 + 0.09645)
=0.4931

Pr(Y<2) = 0.5070

--------

b)

(0.003858)(0.06419 + 0.7061)/(0.5070) = 0.005862

Please explain where each of these numbers have come from because it doesn't look to me like you've calculated the probabilities that are indicated in my first reply.
• Oct 16th 2008, 05:52 AM
chrisc
Pr(y =1) = (1/12)(11/12)^3 = 0.06419
Pr(y=0) = (11/12)^4 = 0.7061

Pr(x=0) = (5/6)^4 = 0.4823
Pr(x=1) = (1/6)(5/6)^3 = 0.09645
Pr(x=2) = (1/6)^2 x (5/6)^2 = 0.01929
Pr(x=3) = (1/6)^3 x (5/6) = 0.003858
Pr(x=4) = (1/6)^4 = 0.0007716

here is a revision. let me know if this is any better.

y1 = (1/12)(0.09645)+(11/144)(0.01929)+(121/1728)(0.003858)+(1331/20736)(0.09645) = 0.01597

y2 = (1)(0.4823)+(11/12)(0.09645)+(121/144)(0.01929)+(1331/1728)(0.003858)+(0.7061)(0.06419) = 0.6352

y1 + y2 = 0.6512

Pr(x=3 | y<2) = 0.004979
• Oct 16th 2008, 07:11 AM
chrisc
or maybe its like this?

y1 = (1/2)(x=1) + (2/4)(x=2) + (3/8)(x=3) + (4/16)(x=4)

the probability of rolling 1 head with 1 six is 1/2
the probability of rolling 1 head with 2 sixes is 2/4 = 1/2
the probability of rolling 1 head with 3 sixes is 3/8
the probability of rolling 1 head with 4 sixes is 4/16 = 1/4

is that right? well going this way, i ended up with

Pr(y<2) = 0.5954

Pr(x=3 | y<2) = 0.003560