leahpiperashe
contestada

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

Respuesta :

Option B

The probability of one of the coins landing on tails and two of them landing on heads is [tex]\frac{3}{8}[/tex]

Solution:

To find:  probability of one of the coins landing on tails and two of them landing on heads

The probability of an event is given as:

[tex]probability =\frac{\text { number of favorable outcomes }}{\text { total number of possible outcomes }}[/tex]

The outcomes are given as:

(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)

Here, total number of possible outcomes = 8

Favorable outcomes = one of the coins landing on tails and two of them landing on heads

Favorable outcomes = (H, H, T) , (H, T, H) , (T, H, H)

Therefore, number of favorable outcome = 3

Thus probability is given as:

[tex]probability = \frac{3}{8}[/tex]

Thus option B is correct

feliciabeal82

Answer:

Step-by-step explanation:

The probability that at least two of the coins will be TAILS is one-half.

Step-by-step explanation:

The probability of an event, E is the ratio of the number of favorable outcomes to the total number of outcomes.

The experiment consisted of tossing three coins together.

The possible outcomes are as follows:

S = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

n (S) = 8

The outcomes where we get at least two Tails are:

s = {(H, T, T), (T, H, T), (T, T, H), (T, T, T)}

n (s) = 4

Compute the probability that at least two of the coins will be TAILS as follows:

                               

Thus, the probability that at least two of the coins will be TAILS is one-half.

Otras preguntas