Answer:
Probability of one of the coin landing on tails and two of them landing on heads is [tex]\frac{3}{8}[/tex]
Step-by-step explanation:
Given:-
Outcomes = (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)
To find:- Probability of 1 coin landing on tails and 2 heads=?
Solution:-
Outcomes = (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)
[tex]\therefore\ n(S)=8[/tex] ----------------(1)
Favourable outcomes = (H,H,T),(H,T,H),(T,H,H)
[tex]\therefore\ n(E)=3[/tex] ------------------(2)
Now using probability formula,
Probability of outcomes = No. of favourable outcomes / Total no. of possible outcomes
Probability of 1 tail and 2 heads = [tex]\frac{favourable\ outcomes}{total\ outcomes}[/tex]
Probability of 1 tail and 2 heads = [tex]\frac{n(E)}{n(S)}[/tex]
Probability of 1 tail and 2 heads = [tex]\frac{3}{8}[/tex] ------(from 1 and 2)
Therefore probability of one of the coin landing on tails and two of them landing on heads is [tex]\frac{3}{8}[/tex]
