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At a game show, there are 8 people (including you and your friend) in the front
row.

The host randomly chooses 3 people from the front row to be contestants.

The order in which they are chosen does not matter.

There are ^8C^3 = 56 total ways to choose the 3 contestants.

What is the probability that you and your friend are both chosen?

A. 2/56
B. 3/56
C. 2/3
D. 6/56

Respuesta :

imlittlestar

Answer:

The correct answer is option D.  6/56

Step-by-step explanation:

It is given that, there are 8 people.

The host randomly chooses 3 people from the front row to be contestants.

To find the required probability

There are 56 ways to choose the 3 contestants.

2 people must be selected. Remaining 1 is selected from 6 people in 6 ways.

Therefore required probability = 6/56

The correct answer is option D.  6/56

vtapert16

Answer:

[tex]\frac{6}{56}[/tex] ~apex

Step-by-step explanation:

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