Question

How do we calculate the sum of the individual probabilities of disjoint events?

Answer 1

P(a) + P(b)

Answer 2

what does the a and b stand for?

Answer 3

Usually the question would ask what is the probability of a or b happening. Could stand for two different objects of the same kind. For example, suppose you have a fruit basket of 5 apples and 7 oranges. And then you choose 1 piece of fruit. What is the probability that you choose an apple or an orange.

Then you would compute the following

\[P(\text{a or b}) = \frac{\text{total apples + total oranges}}{\text{total fruit}}\]

Answer 4

Actually that's a bad example.

Answer 5

lol

Answer 6

Okay so maybe you have some pears as well. Like 4 pears

Answer 7

so then it would be

\[\frac{5 + 7}{16}\]

or \[\frac{5}{16} + \frac{7}{16}\]

Answer 8

i think i get it, like the two combinations the 2 fruits and the total of the 2 on the bottom ?

Answer 9

or \[\frac{3}{4}\]

Answer 10

but what do the a and b stand for?

Answer 11

is it like the 2 fruits?

Answer 12

They stand for subsets of an entire set of objects.

Answer 13

All the fruit in the basket represents the entire set. a represents all the apples in the set. b represents all the oranges in the set.

Answer 14

oh,

Answer 15

Say that set S represents the total fruit in the basket and

a = all the apples
b = all the oranges
c = all the pears

You can represent the set in the following manner:

\[s = \brace{a_1, a_2, a_3, a_4, a_5, b_1, b_2, b_3, b_4, b_5, b_6, b_7, c_1, c_2, c_3, c_4}\]

Answer 16

a = {a1, a2, a3, a4, a5}
b = {b1, b2, b3, b4, b5, b6, b7}
c = {c1, c2, c3, c4}

Answer 17

i get it..... lol, thanks!