Answer:
a) 1/11
b) 25/66
c) 41/66
Explanation:
The following is the number of each type of can
Cola (C) = 6
Soda (S) = 5
Fizz (F) = 1
Total number of cans = 12
Since the sampling is done without replacement, the probability will be different for different draws
Let P(C₁) = Probability of drawing a cola on first draw
P(C₁) = 6/12
P(C₂|C₁) = Probability of cola on second draw given that the first draw was a cola = 5/11 (11 total cans left for second draw and only 5 cans of cola)
The probabilities for the other two types of cans can be calculated in the same way
P(S₁) = 5/12
P(S₂|S₁) = 4/11
P(F₁) = 1/12
P(F₂|F₁) = 0/11 = 0 (since there is only one can of Fizz the probability of drawing a second can of Fizz is 0
a)
In two draws what is the probability that one can is C and other is F
There are two ways in which this can occur - C₁ F₂ and F₁C₂
So the combined probability = sum of these probabilities for both possibilities
P(one C and one F) = P(C₁F₂) + P(F₁ C₂)
P(C₁F₂) = P(C₁) · P(F₂|C₁) = 6/12 · 1/11 = 1/2 · 1/11 = 1/22
P(F₁C₂) = P(F₁) · P(C₂|F₁) = 1/12 · 6/11 = 1/12 · 6/11 = 1/22
So P(C₁F₂ or F₁C₂) = 1/22 + 1/22 = 2/22 = 1/11
b)
P(both cans having same contents).
This can be represented as
P(C₁C₂ or S₁S₂ or F₁F₂)
= P(C₁C₂) + P(S₁S₂) + P(F₁F₂)
= P(C₁) x P(C₂|C₁) + P(S₁) x P(S₂|S₁) + P(F₁) x P(F₂|F1)
= 6/12 x 5/11 + 5/12 x 4/11 + 1/12 x 0
= 50/132
= 25/66
c)
Probability that the two cans will differ is the complement of the event the the two cans have the same contents
P(complement of event E) = 1 - P(event E)
P(can contents differ) = 1 - P(can contents are the same)
= 1 - 25/66
= 41/66
I hope I got it right, please let me know .Thanks