Conditional Probability

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
1. A coin is tossed 3 times. De ne the events A and B as follows:
A = \at least 2 heads are obtained", B = \all 3 tosses are the same".
(a) Calculate P(A).
(b) Calculate P(A / B).
(c) Are A and B independent events? Explain.

I think for a the answer is P(A) = 5/6, which is 0.83

for, b, and c Im not sure how to do it.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,184
quaidy4 said:
1. A coin is tossed 3 times. De ne the events A and B as follows:
A = \at least 2 heads are obtained", B = \all 3 tosses are the same".
(a) Calculate P(A).
(b) Calculate P(A / B).
(c) Are A and B independent events? Explain.

I think for a the answer is P(A) = 5/6, which is 0.83

for, b, and c Im not sure how to do it.
First draw a tree diagram - and count the events. There are total eight events.

Now continue.....
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, quaidy4!

Here are the first two parts . . .


\(\displaystyle \text{1. A coin is tossed 3 times.}\)

\(\displaystyle \text{Define the events }A\text{ and }B\text{ as follows: }\;\begin{Bmatrix} A:& \text{at least 2 heads are tossed} \\B: & \text{all 3 tosses are the same} \end{Bmatrix}\)

\(\displaystyle \text{(a) Calculate }P(A)\)

\(\displaystyle \text{There are }eight\text{ possible outcomes:}\)

. . \(\displaystyle \begin{array}{cccccc} HHH & * \\ HHT & * \\ HTH & * \\ HTT \\ THH & * \\ THT \\ TTH \\ TTT \end{array}\)


\(\displaystyle \text{In }f\!our\text{ cases there are at least 2 heads.}\)

\(\displaystyle \text{Therefore: }\:p(A) \;=\;\frac{4}{8} \:=\:\frac{1}{2}\)




\(\displaystyle \text{(b) Calculate }P(A|B)\)

\(\displaystyle \text{"The probabiity that there are at least 2 heads, given that all three tosses are the same."}\)


\(\displaystyle \text{If all three tosses are the same, there are only }two\text{ cases in the sanple space: }\:HHH,\,TTT\)

\(\displaystyle \text{And in }one\text{ of them, there are at least 2 heads: }\:HHH\)

\(\displaystyle \text{Therefore: }\;P(A|B) \;=\;\frac{1}{2}\)

 

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
oh okay, thank you both for all your help! I don't know what I was thinking that there was only 6 possible outcomes, my bad.


Also, for c I believe the answer is yes A and B are independent because P(A) = 1/2 and P(A/B) = 1/2 they both equal the same.
 

saravananbs

New member
Joined
Jan 6, 2011
Messages
6
if A and B are independent then P(AB)=P(A)*P(B) otherwise they are not.
 

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
oh okay, so A and B are not independent.
 
Top