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Combined Events Simplified Revision Notes

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Combined Events

Combined Events

Union of Two Events (A or B)

The probability that either event AA or event B B (or both) occur is given by:

P(A or B)=P(A)+P(B)P(A and B) P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
infoNote

Example: Calculate the probability of drawing a card that is either a heart or an Ace from a deck of cards:

Step 1: Calculate the probability of drawing a heart from a deck of cards.

There are 52 cards in a deck, and 13 of them are hearts.

P(Heart)=1352=14P(\text{Heart}) = \frac{13}{52} = \frac{1}{4}

Step 2: Calculate the probability of drawing an Ace from a deck of cards.

There are 52 cards in a deck, and 4 of them are Aces.

P(Ace)=452=113P(\text{Ace}) = \frac{4}{52} = \frac{1}{13}

Step 3: Calculate the probability of drawing the Ace of Hearts from a deck of cards.

There are 52 cards in a deck, and only one of them is the Ace of Hearts.

P(Ace of Hearts)=152P(\text{Ace of Hearts}) = \frac{1}{52}


Step 4: Using the formula, calculate the probability of drawing a card that is either a heart or an Ace from a deck of cards:

Use this formula:

P(A or B)=P(A)+P(B)P(A and B) P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)

Insert the values from the previous probability calculations.

P(Heart or Ace)=14+113152P(\text{Heart or Ace}) = \frac{1}{4} + \frac{1}{13} - \frac{1}{52} =1352+452152= \frac{13}{52} + \frac{4}{52} - \frac{1}{52} =1652=413= \frac{16}{52} = \frac{4}{13}

Intersection of Two Events (A and B)

The probability that both event AA and event BB occur is:

P(A and B)=P(A)×P(BA)(if A and B are dependent) P(A \text{ and } B) = P(A) \times P(B \mid A) \quad \text{(if A and B are dependent)}
infoNote

Example: Calculate the probability of drawing two Aces in a row without replacement from a deck of cards:

Step 1: Calculate the probability of drawing the first Ace.

There are 52 cards in a deck, and 4 of them are Aces.

Therefore the probability of drawing the first Ace is:

452=113 \frac{4}{52} = \frac{1}{13}

Step 2: Calculate the probability of drawing the second Ace.

Since one Ace has already been picked up there are now 51 cards in the deck, with 3 Aces.

Therefore the probability of drawing the second Ace is:

351=117 \frac{3}{51} = \frac{1}{17}

Step 3: Using the formula calculate the probability of drawing two Aces in a row without replacement from a deck of cards.

P(Two Aces)=113×117=1221P(\text{Two Aces}) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}
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