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Probability (Higher) - Conditional Probability

Conditional Probability Example | Harder Problems

Conditional Probability Example

This is when the probability of an event is influenced by a previous event.

Example: Two discs are selected from a bag of 6 blue and 4 green discs, without replacing the first disc.

a) What is the probability of getting two discs of the same colour?

P (B) = 6/10 for the first disc.

P (B) = 5/9 for the second disc because there are only 5 blue discs left and 9 discs in the bag.

So P (BB) = 6/10 x 5/9 = 30 /90

	P (GG) = 4/10 x 3/9 =12/90

	P (same colour)	= 30/90 +12/90
		= 42/90
		= 7/15
b) What is the probability of getting two different coloured discs?

	P (BG)	= 6/10 x 4⁄9 = 24⁄90
	P (GB)	= 4/10 x 6⁄9 = 24⁄90
	P (different colour)	= 24⁄90 + 24⁄90
		= 48⁄90 = 16⁄30 = 8/15

Harder Problems

'At least' problems
Example1: in the Conditional Probabilities problem above, what is the probability of selecting at least one green disc?

In this case the only outcome we are not interested in is two blue discs. So we can write:

This can be written as a general rule:

P (at least one) = 1 – P (none)

Other problems
Example 2: Boris and John play three sets of tennis. The probability of Boris winning the first set is 0.4.

If Boris wins a set the probability of his winning the next set is 0.7.

If John wins a set the probability of his winning the next set is 0.8.

a) Calculate the probability that John wins all three sets.

P (John wins 3 sets)	= 0.4 x 0.7 x 0.7
	= 0.196

b) Calculate the probability that Boris wins at least one set.

Example 3: when I go to work, I have to pass two sets of traffic lights, A and B.

The probability that I have to stop at A is 0.4.

If I have to stop at A, the probability that I have to stop at B is 0.8.

If I do not stop at A, the probability that I have to stop at B is 0.3.

What is the probability that I have to stop at both A and B?

Calculate the probability that I have to stop at least once.

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