Latest revision as of 11:34, 27 January 2013

Conditional Probability Problem

You roll a 6 faced fair die then take the number you rolled and flip a fair coin that number of times. (ex: If you roll 3, flip the coin 3 times.) P(A)= The probability of rolling a 4 P(B)= The probability of tossing heads and tails in equal number of times

Find P(A|B).

Solution

Finding the Probability of A.

P(A)=1/6, the probability of rolling a 4 out of 6 outcomes.

Finding the Probability of B

There is a 0 chance of B if die was rolled a odd number. If rolled a 2, there are 2 sets (2*1). If rolled a 4, there are 6 sets (3*2). If rolled a 6, there are 20 sets(5*4). |B|=2+6+20=28 B={(2,H,T),(2,T,H)

  (4,H,H,T,T),(4,H,T,H,T),(4,H,T,T,H),(4,T,H,H,T),(4,T,H,T,H),(4,T,T,H,H)...}

P(B)=28/126

\Omega={(1,H),(1,T)

       (2,H,H),(2,H,T),(2,T,H),(2,T,T)...}

Finding P[AnB]

A is a subset of B. It's the case if rolled a 4, there are 6 sets. therefore P[AnB]=6/126

P[A|B]=P[AnB]/P[B]=(6/126)/(28/126)=3/14

Comments

Back to first bonus point opportunity, ECE302 Spring 2013