# Thread: help/confirmation on multi-stage question

1. ## 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?

2. 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).

3. 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

4. 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.

5. 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

6. 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