Answer:
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.
Step-by-step explanation:
Hello!
You have the following possible pizza orders:
Onion ⇒ P(on)= 0.36
Sausage ⇒ P(sa)= 0.35
Onions and Sausages ⇒ P(on∪sa)= 0.66
The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.
If two events are not mutually exclusive, you know that:
P(A∪B)= P(A)+P(B)-P(A∩B)
Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:
P(on∪sa)= P(on)+P(sa)-P(on∩sa)
P(on∪sa)+P(on∩sa)= P(on)+P(sa)
P(on∩sa)= P(on)+P(sa)-P(on∪sa)
P(on∩sa)= 0.36+0.35-0.66= 0.05
I hope it helps!
3.50 x 3 = 10.50
6.85 x 2 = 13.70
10.50 + 13.70 = 24.20
Step-by-step explanation:
- Number of red = 2
- Number of blue = 5
- Number of green = 3
- total number of marbles = 10
<h3>
probability of not choosing a red marble = 1--choosing a red marble.</h3>
<u>Because</u><u> </u><u>probability</u><u> </u><u>is</u><u> </u><u>always</u><u> </u><u>one</u><u>(</u><u>1</u><u>)</u><u>.</u>
<em>Probability</em><em> </em><em>=</em>
<em>
</em>
<em>
</em>
<em>
</em>
<em>Is</em><em> </em><em>the</em><em> </em><em>probability</em><em> </em><em>of</em><em> </em><em>not</em><em> </em><em>choosing</em><em> </em><em>a</em><em> </em><em>red</em><em> </em><em>marble</em><em>.</em>
Plugging in the value "
" in the above expression, we have
<h3><u>Note</u>:-</h3>
P = Parentheses
E = Exponents
M = Multiplication
D = Division
A = Addition
S = Subtraction
