Home >>STATISTICS, Section 1, probability 3
Non-Mutually Exclusive Events |
Non-Mutually Exclusive Events ('Compatible')
Two or more 'non-mutually exclusive' events can occur at the same time.
In every case one event does not prevent the other happening.
The probability of either one OR both events occuring is:
where,
P(A ∪ B) is the probability of event A OR event B happening at the same time
P(A ∩ B) is the probability of event A AND event B happening at the same time
Example #1
What is the probability of getting a black card OR an Ace by drawing one card from a 52 deck.
P(Black) = 26/52 = 1/2
P(Ace) = 4/52 = 1/13
P(Black ∩ Ace) = 1/2 x 1/13 = 1/26
using, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(Black ∪ Ace) = P(Black) + P(Ace) - P(Black ∩ Ace)
P(Black ∪ Ace) = 26/52 + 4/52 - 1/26
P(Black ∪ Ace) = 30/52 - 2/52 = 28/52 = 7/13 |
A card can either be Black or Ace or both (i.e. a Black Ace).
So that's why we need to subtract the probability of a card being both Black AND Ace .
This card has already been accounted for in the probability of the card being Black AND the probability of the card being Ace.
Example #2
For a 6 sided die, what is the probability of obtaining an even number OR a number greater than 3?
Both events, A and B, occur when either a 4 or 6 are thrown:
A - the event of getting an even number (2, 4, 6)
B - the event of getting a number greater than 3 (4, 5, 6)
P(A) = 3/6 = 1/2
P(B) = 3/6 = 1/2
P(A ∩ B) = 1/4
using, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 1/2 + 1/2 - 1/4 = 3/4 |
Example #3
For a set of dominoes, what is the probability of choosing one domino and it containg a 5 or any double?
P(5) = 7/28 = 1/4 ( 5,0 5,1 5,2 5,3 5,4 5,5 5,6 )
P(double) = 7/28 = 1/4 ( 0,0 1,1 2,2 3,3 4,4 5,5 6,6 )
P(5 ∩ double) = 1/16
using, P(A) ∪ P(B) = P(A) + P(B) - P(A ∩ B)
P(5) ∪ P(double) = P(5) + P(double) - P(5 ∩ double)
P(5) ∪ P(double) = 1/4 + 1/4 - 1/16 = 4/16 + 4/16 - 1/16 = 7/16 |
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