Question

One ball is selected at random from each box.
Find the probablity that the three selected balls
consist of 1 red and 2 green balls

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{ One ball is selected at random from each box.}\hspace{.33em}\\~\\
& \normalsize \text{ Find the probablity that the three selected balls }\hspace{.33em}\\~\\
& \normalsize \text{ consist of 1 red and 2 green balls }\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

I'm very sorry, I don't know your answer, I'm not good with probability

Answer 3

answer given is 3/8

Answer 4

So you can have 1 red and 2 green in the following ways:

A=rgg
B=ggr
C=grg

Since each case in mutually exclusive the probability is
P(A)+P(B)+P(C)

For Event A
$$
P(A)=3/4*2/4*2/3=\\
$$
For Event B
$$
P(B)=1/4*2/4*1/3\\
$$
For Event C
$$
P(C)=1/4*2/4*2/3\\
$$

So total probability, p is
$$
p=3/4*2/4*2/3+1/4*2/4*1/3+1/4*2/4*2/3=3/8
$$

Does this make sense?

Answer 5

can i help

Answer 6

nvm

Answer 7

yep thnks

Answer 8

yw