Question

PLEASE HURRY THANK YOU

Answer 1

Answer:

Probability of one of the coin landing on tails and two of them landing on heads is $\frac{3}{8}$

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)

$\therefore\ n(S)=8$     ----------------(1)

Favourable outcomes = (H,H,T),(H,T,H),(T,H,H)

$\therefore\ n(E)=3$    ------------------(2)

Now using probability formula,

Probability of outcomes = No. of favourable outcomes / Total no. of possible outcomes

Probability of 1 tail and 2 heads = $\frac{favourable\ outcomes}{total\ outcomes}$

Probability of 1 tail and 2 heads = $\frac{n(E)}{n(S)}$

Probability of 1 tail and 2 heads = $\frac{3}{8}$    ------(from 1 and 2)

Therefore probability of one of the coin landing on tails and two of them landing on heads is $\frac{3}{8}$

Answer 2

Answer:

the answer is B  3/8

Step-by-step explanation: