A box contains 5 red bowls, 6 yellow bowls and 3 white bowls. If 4 bowls are randomly drawn from the box at the same time,
(a)find the probability that exactly 2 red bowls are drawn;
(b)find the probability that at least 2 red bowls are drawn.
My work:
(a) (5/14)^2(6/14)(3/14)+(5/14)^2(6/14)^2+(5/14)^2(3/14)^2=225/19208
(b) 225/19208+(5/14)^3(6/14)+(5/14)^3(3/14)+(5/14)^4=275/4802
To solve this problem, we need to first determine the total number of ways to draw 4 bowls from the box. Then, we will calculate the number of favorable outcomes for each scenario (exactly 2 red bowls and at least 2 red bowls).
Let's break down the solution step by step:
Step 1: Calculate the total number of ways to draw 4 bowls from the box.
Since we are drawing 4 bowls from a total of 14 bowls (5 red + 6 yellow + 3 white), we can use combinations to calculate the total number of ways to draw. The formula for combinations is nCr, where n is the total number of items and r is the number of items chosen.
So, the total number of ways to draw 4 bowls is given by:
Total ways = 14C4 = 14! / (4! * (14 - 4)!) = 14! / (4! * 10!) = (14 * 13 * 12 * 11) / (4 * 3 * 2 * 1) = 1001.
Step 2: Calculate the number of favorable outcomes for each scenario.
(a) Find the probability that exactly 2 red bowls are drawn.
In this case, we need to consider three different possibilities:
(i) Two red bowls, one yellow bowl, and one white bowl:
To calculate the number of ways to choose 2 red bowls, 1 yellow bowl, and 1 white bowl, we have 5C2 * 6C1 * 3C1. Applying the combination formula for each color, we get:
Number of ways = (5! / (2! * (5 - 2)!) ) * (6) * (3) = 10 * 6 * 3 = 180.
(ii) Two red bowls and two yellow bowls:
Number of ways = 5C2 * 6C2 = (5! / (2! * (5 - 2)!) ) * (6! / (2! * (6 - 2)!) ) = 10 * 15 = 150.
(iii) Two red bowls and two white bowls:
Number of ways = 5C2 * 3C2 = (5! / (2! * (5 - 2)!) ) * (3! / (2! * (3 - 2)!) ) = 10 * 3 = 30.
Summing up the possibilities (i), (ii), and (iii) gives the number of favorable outcomes for exactly 2 red bowls: 180 + 150 + 30 = 360.
(b) Find the probability that at least 2 red bowls are drawn.
To find the probability of drawing at least 2 red bowls, we need to calculate the number of favorable outcomes for the following scenarios:
(i) Exactly 2 red bowls (already calculated in part a): 360.
(ii) Exactly 3 red bowls:
Number of ways = 5C3 * 6C1 * 3C0 = (5! / (3! * (5 - 3)!) ) * 6 * 1 = 10 * 6 = 60.
(iii) All 4 red bowls:
Number of ways = 5C4 * 6C0 * 3C0 = (5! / (4! * (5 - 4)!) ) * 1 * 1 = 5.
Summing up the possibilities (i), (ii), and (iii) gives the number of favorable outcomes for at least 2 red bowls: 360 + 60 + 5 = 425.
Step 3: Calculate the probability for each scenario.
(a) Probability of exactly 2 red bowls:
P(exactly 2 red bowls) = Number of favorable outcomes / Total number of outcomes = 360 / 1001 = 360/1001.
(b) Probability of at least 2 red bowls:
P(at least 2 red bowls) = Number of favorable outcomes / Total number of outcomes = 425 / 1001 = 425/1001.
Therefore, the final answers are:
(a) The probability that exactly 2 red bowls are drawn is 360/1001.
(b) The probability that at least 2 red bowls are drawn is 425/1001.