Question

PLEASE HURRY THANK YOU

The tree diagram below shows all of the possible outcomes for flipping three coins.

A tree diagram has 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).

What is the probability of one of the coins landing on tails and two of them landing on heads?
1/4
3/8
1/2
3/4

Answer 1

the answer is B 3/8

Explanation:

Answer 2

Probability of one of the coin landing on tails and two of them landing on heads is
`(3)/(8)`

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 =
`(favourable\ outcomes)/(total\ outcomes)`

Probability of 1 tail and 2 heads =
`(n(E))/(n(S))`

Probability of 1 tail and 2 heads =
`(3)/(8)` ------(from 1 and 2)

Therefore probability of one of the coin landing on tails and two of them landing on heads is
`(3)/(8)`