Question

A bag contains two red balls and five blue balls. You draw out a ball, record its color, replace it, and then draw out a second ball, record its color, etc., repeating this four times. What is the probability that you draw out a red ball AT LEAST once?

Answer 1

B) 0.49

Explanation:

Answer 2

0.74

Explanation:

Drawing at least 1 red ball is the same as: draw 4 red balls or draw 3 red balls or draw 2 red balls or draw 1 red ball. You can calculate all these probabilities and sum them as they are independent:

P (at least 1 red) = P (1 red) + P (2 red) + P (3 red) + P (4 red)

However you can think about the inverse success "not drawing any red" and say that:

P (at least 1 red) = 1 - P (not any red)

This is because the fact of gettin at least 1 red is the complement of getting no red ball. What happens if you do not get at east 1 red ball? Then, you get not any red ball (0 red balls).

Now we need to see the prob of getting no red ball at all, which is the prob of getting a blue ball in every draw. As we have replacing the probabilities are independent. The prob of getting 1 blue ball is:

P(1 blue) = 5/7 as there are 5 blue balls over a total 7 balls.

Now, the prob of getting another blue ball, given that you got a blue on the first is simply the product:

P(blue blue) = 5/7 * 5/7 = 25/49

So, for the third blue ball:

P(blue blue blue) = 5/7 * 5/7 * 5/7 = 125 / 343

And the forth blue ball:

P (blue blue blue blue) = 5/7 * 5/7 * 5/7 * 5/7 = 625 / 2,401 = 0.26

So, the probability of getting no red balls is 0.26. Then, the probability of getting some red ball is:

P (red) = 1 - 0.26 = 0.74