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Probability: AND & OR rule

The ‘OR’ rule

The ‘OR’ rule states P(A\text{ or } B)=P(A)+P(B)

This makes sense since the number of successes(NoS) = (NOS of A) + (NoS of B)

The number of outcomes has not changed.

So P(A\mbox{ or } B)=\frac{\mbox{(NOS of A) + (NoS of B)}}{\mbox{possible outcomes}}=\frac{\mbox{(NOS of A)}}{\mbox{possible outcomes}}+\frac{\mbox{(NOS of A)}}{\mbox{possible outcomes}}

\implies P(A\mbox{ or } B)=P(A)+P(B)


A bag contains a red ball, 5 blue ball and 4 green balls.

1) P(red)

2) P(blue)

3) P(green)

4) P(red or green)

5) P(red or green or blue)


1) P(red)=\frac{1}{10}

2) P(blue)=\frac{5}{10}

3) P(green)=\frac{4}{10}

4) P(\mbox{red or green})=P(red)+P(green)=\frac{1}{10}+\frac{4}{10}=\frac{5}{10}

5) P(\mbox{red or green or blue})=P(red)+P(green)+(blue)=\frac{1}{10}+\frac{4}{10}+\frac{5}{10}=1

The ‘AND’ rule

Here we are considering the probability of something happening and then something else happening.

Let’s consider a die rolled twice.

What is the P(two sixes)?

Well we will get a ‘6’ \frac{1}{6} of the time.  Of these times we will get another ‘6’ \frac{1}{6} of the time.

So P(\mbox{two sixes})=\frac{1}{6}\text { of }\frac{1}{6}=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}

So P(\mbox{A and B})=P(A)\times P(B)


A bag contains 4 red(R) balls and 3 black(B) balls.  I take out a ball, look at it and then put it back.  I then take out another ball and record its colour.  What is the:

  1. P(R,R) \rightarrow means getting two reds.
  2. P(R then B)
  3. P(R,B) \rightarrow means getting a red and blue in any order. i.e. red then blue or blue then red





P(R,R)=\frac{4}{7} \times\frac{4}{7} =\frac{16}{49}

P(\mbox{R then B})=\frac{4}{7}\times \frac{3}{7}=\frac{12}{49}

P(\mbox{R, B})=P(\mbox{R then B}) + P(\mbox{B then R})

P(\mbox{B then R})=\frac{3}{7}\times \frac{4}{7}=\frac{12}{49}



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