Answer:
the probability of obtaining 2 heads and 1 tail in any order is higher than the probability of obtaining a head, than a head then a tail
Step-by-step explanation:
since the probability (p) of obtaining 2 head and 1 tail in any order is
P1 = p of obtaining ( H , H ,T) + p of obtaining ( H , T ,H ) + p of obtaining ( T , H ,H ) = 3*p of obtaining ( H , T ,H)
assuming a fair coin then p heads = p tails = 0.5
thus since each flip is independent from the others
p( H , H ,T)=p( H , T ,H )= p( T , H ,H )= P=0.5*0.5*0.5= 1/8
thus P1 =3*1/8=3/8
while the probability of obtaining a head, than a head then a tail is
P2= p of obtaining ( H , T ,H)= 1/8
then P1=3/8 >P2=1(8
therefore the probability of obtaining 2 heads and 1 tail in any order is higher than the probability of obtaining a head, than a head then a tail
- Original point
- x-Coordinate of point P
- Resulting point