Question 349911
I'll do 4 of the problems.
.
.
a) P(T)=P(H)=1/2, it's independent of previous tosses if the coin is fair.
.
.
.
b) P(6)=1/6
.
.
.
c) P(Ace)=4/52=1/13
P(King of Hearts)=1/52
P(Spade)=4/52=1/13
P(Ace or K Hearts or Spade)=P(Ace)+P(K Hearts)+P(Spade)-P(Ace of Spades)
So we don't double count the Ace of Spades as an Ace and a Spade.
P(Ace of Spades)=1/52
 P(Ace or K Hearts or Spade)=4/52+1/52+13/52-1/52=17/52
.
.
.
d) P(B,NB,NB)=0.3(0.7)(0.7)
P(NB,B,NB)=0.7(0.3)(0.7)
P(NB,NB,B)=0.7(0.7)(0.3)
P(1 B)=3(0.3)(0.7)(0.7)=0.441