Question 998790
 
Problem:
 
Two biased coins are tossed simultaneously.
Coin A has probability of 0.3 for heads, and
coin B has probability of 0.4 for heads.
Find
(a) two heads
(b) coin A shows head
(c) one head
 
Solution:
 
Define event Ha=coin A shows heads, Hb=coin B shows heads.
 
(a) Two heads
The two coins are independent of each other, so probability of both events (heads) happening is the product of the individual probabilities.
Probability of two heads 
= P(Ha and Hb)
= P(Ha)*P(Hb)
= 0.3*0.4
= 0.12
 
(b) Coin A shows head
This means that the outcome of coin B is immaterial, so
P(Ha)=0.3
 
(c) one head 
This means exactly one head, which is either 
event Ha and ~Hb  
or
event ~Ha and Hb.
The two events are independent and mutually exclusive, so we can add the two probabilities.
P(Ha and ~Hb)=0.3*(1-0.4)=0.18
P(~Ha and Hb)=(1-0.3)*0.4=0.28
Probability of showing exactly one head
= 0.18+0.28
= 0.46