Question No. 1 [6 Marks] Suppose you have a box with 3 blue marbles, 4 red marbles and 5 green marbles. You are required to pull out 7 marbles randomly from the box. Answer the following questions. a) Find the probability of drawing at least 2 green marbles. b) Find the probability of drawing at most 3 green marbles. c) Find the probability of drawing exactly 4 green marbles. d) Find the probability of getting atleast 1 blue and atleast 1 green marble. e) Find the probability of getting atleast 1 blue and atleast 1 red marble. f) Find the probability of getting atleast 1 blue and atleast 1 green marble and atleast 1 red marble.

a) To find the probability of drawing at least 2 green marbles, we first need to calculate the total number of ways to draw 7 marbles from 12 (3 blue, 4 red, 5 green). This can be done using the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, and r is the number of items to choose. So, the total number of ways to draw 7 marbles from 12 is C(12, 7) = 792.

Next, we calculate the number of ways to draw at least 2 green marbles. This is the sum of the ways to draw 2, 3, 4, 5 green marbles and the rest from the other colors. So, we have:

• 2 green: C(5, 2) * C(7, 5) = 10 * 21 = 210
• 3 green: C(5, 3) * C(7, 4) = 10 * 35 = 350
• 4 green: C(5, 4) * C(7, 3) = 5 * 35 = 175
• 5 green: C(5, 5) * C(7, 2) = 1 * 21 = 21

Adding these up gives 210 + 350 + 175 + 21 = 756 ways to draw at least 2 green marbles.

So, the probability of drawing at least 2 green marbles is 756 / 792 = 0.9545 or 95.45%.

b) The probability of drawing at most 3 green marbles is the sum of the probabilities of drawing 0, 1, 2, or 3 green marbles. We calculate these probabilities in a similar way as in part a), and then subtract the result from 1 (since the sum of the probabilities of all possible outcomes is 1).

c) The probability of drawing exactly 4 green marbles is calculated as the number of ways to draw 4 green marbles and 3 from the other colors, divided by the total number of ways to draw 7 marbles. So, we have C(5, 4) * C(7, 3) / C(12, 7) = 175 / 792 = 0.221 or 22.1%.

d) The probability of getting at least 1 blue and at least 1 green marble is calculated as 1 minus the probability of not getting any blue or green marbles. The number of ways to draw 7 marbles without any blue or green is C(4, 7), which is 0 since we can't draw 7 marbles from 4. So, the probability is 1 - 0 = 1 or 100%.

e) The probability of getting at least 1 blue and at least 1 red marble is calculated in a similar way as in part d), but now we consider the number of ways to draw 7 marbles without any blue or red, which is C(5, 7) = 0. So, the probability is also 1 or 100%.

f) The probability of getting at least 1 blue, at least 1 green, and at least 1 red marble is calculated as 1 minus the probability of not getting any of these colors. The number of ways to draw 7 marbles without any blue, green, or red is 0, so the probability is 1 - 0 = 1 or 100%.