You randomly draw twice from this deck of cards

Mathematics

1 answer:

Serggg [28]2 years ago

3 0

The probability of not drawing C in neither draw is P = 0.5

<h3>How to get the probability?</h3>

All the cards have the same probability of being drawn, in this case, our set of cards is {F, D, C, G}

The probability of not drawing C is equal to the probability of drawing F, D or G. So we have 3 options out of 4, then the probability is:

p = 3/4.

Now we draw another, this time there are 3 cards, one of these is C, and the other two cards are not C. Then the probability of not drawing C again is equal to 2 over 3.

q = 2/3.

The joint probability (for both of these events to happen) is equal to the product of the individual probabilities:

P = p*q = (3/4)*(2/3) = 0.5

If you want to learn more about probability, you can read:

brainly.com/question/251701

You might be interested in

What are the solutions of the following system

noname [10]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}10x^2-y=48\\2y=16x^2+48&\text{subtracy}\ 16x^2\ \text{from both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}10x^2-y=48\\-16x^2+2y=48&\text{divide both sides by 2}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}10x^2-y=48\\-8x^2+y=24\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2x^2=72\qquad\text{divide both sides by 2}\\.\qquad x^2=36\to x=\pm\sqrt{36}\\.\qquad x=\pm6\\\\\text{Put the values of}\ x\ \text{to the second equation}$