A small deck of 5 cards consists of 3 red Jards and 2 green cards. Draw 7 times with replacement.
Find the probability that a red card is drawn exactly 3 times.

Approximately .

Step-by-step explanation:

The cards were drawn with replacement. Therefore, the probability of drawing a red card will be at every one of the seven draws. The probability of not drawing a red card would be .

One way to get three red cards out of the seven draws would be . That corresponds to a probability of .

However, there are many other ways to get three red cards during the seven draws. For example:

There are such ways in total to choose three draws out of seven draws.

These ways are mutually-exclusive. Each of these ways has a probability of . The probability for getting a red card on exactly three out of the seven draws would be:

.

15

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?