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N a competition for children at a local primary school, a container is filled with six cans of

Cola, five cans of Soda and one can of Fizz. The cans are identical in all respects except for

their contents. If a child is required to draw two cans without replacement from the

container, what is the probability that:

a) One can filled with Cola and one can filled with Fizz will be drawn

b) The cans will have the same contents?

c)The cans will differ with respect to their contents?

1 Answer

6 votes

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


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