Answer:
Step-by-step explanation:
Hello!
Given the probabilities:
P(A₁)= 0.35
P(A₂)= 0.50
P(A₁∩A₂)= 0
P(BIA₁)= 0.20
P(BIA₂)= 0.05
a)
Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)
Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.
b)
Considering that 
you can clear the intersection from the formula 
and apply it for the given events:
c)
The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:
P(B)= (A₁∩B) + P(A₂∩B)= 0.07 + 0.025= 0.095
d)
The Bayes' theorem states that:
Then:
I hope it helps!
Answer:
Eli=180 Freda=260 and Geoff=360
Answer:
11/15
Step-by-step explanation:
Divide both the numerator and denominator by the GCD
55 ÷ 5
75 ÷ 5
=11/15
:)
Step-by-step explanation:
The algebraic expression is
20
----
c
or
20/c
Answer:
Trial- 2 shows the conservation of momentum in a closed system.
Step-by-step explanation:
Given: Mass of balls are 
Conservation of momentum in a closed system occurs when momentum before collision is equal to momentum after collision.
- Let initial velocity of ball
- Initial velocity of ball
- Final velocity of ball
- Final velocity of ball
- Momentum before collision
- Momentum after collision
Now, According to conservation of momentum.
Momentum before collision = Momentum after collision
We will plug each trial to this equation.
Trial 1
Trial 2
Trial 3
Trial 4
We can see only Trial 2 satisfies the princple of conservation of momentum. That is momentum before collison should equal to momentum after collision.
