Answer part 1.
P(Shaun loses both) = (1-3/8)(1-5/7) = (5/8)(2/7) = 10/56
Step-by-step explanation part 1.
P(Shaun wins over Mike) = 3/8
P(Shawn loses to Mike) = 1 - 3/8
P(Shawn wins over Tim) = 5/7
P(Shawn loses to Tim) = 1 - 5/7
Events are independent so P(A and B) = P(A)P(B)
Answer part 2:
Scenario 1, revised to make it solvable.
Event A is the set of all outcomes where a child likes chocolate cupcakes, P(A) = 70%.
Event B for lemon cupcakes with P(B) = 30%.
P(A ∩ B) = 25%.
Test for Independence:
P(A)P(B) = 0.7×0.3 = 0.21 < 25% = P(A ∩ B)
The events are not independent.
P(B|A) = P(A ∩ B) / P(A) = 25%/70% = 36% > P(B)
P(A|B) = P(A ∩ B) / P(B) = 25%/30% = 83% > P(A)
Scenario 2, revised:
Event B is "a player is selected for offense", P(B) = 60%, and event A is "a player is selected for defense", P(A) = 40%. P(A ∩ B) = 24%.
Test for Independence:
P(A)P(B) = 0.6×0.4 = 24% = P(A ∩ B).
The events are independent.
P(B|A) = P(A ∩ B) / P(A) = 24%/60% = 40% = P(B)
P(A|B) = P(A ∩ B) / P(B) = 24%/40% = 60% = P(A)
Scenario 3, revised:
A is the event that a person chooses mud run. Estimate of P(A) from 120 trials is 40/120 = 33%. B is the event that a person chooses river rafting. Estimate of P(B) is 60/120 = 50%. Estimate P(A ∩ B) = 30/120 = 25%.
Test for Independence:
P(A)P(B) = (1/3)(1/2) = 1/6 = 17% < 25% = P(A ∩ B).
The events are not independent.
P(B|A) = P(A ∩ B) / P(A) = 25%/33% = 75% > 50% = P(B)
P(A|B) = P(A ∩ B) / P(B) = 25%/50% = 50% > 33% = P(A)
This problem is seriously garbled.
Problem as stated in photo. (Thanks Google Lens for converting to text. Only a few corrections were needed.)
Analyze the conditional probability P(B|A), for each scenario given in the first column and thus classify them as dependent and independent events under 2 column headings.
Scenario 1: 'A' be the event that 70% of the children like chocolate cupcakes and 'B' be the event that 25% like lemon cupcakes. 30% of children like both.
Scenario 2: 'B' be the event that 60% of the players are selected for offensive side and 'A' be the event that 40% are selected for defensive side. 24% are selected as reserved players for both sides.
Scenario 3 : Consider a group of 120 people. 'A' be the event that 40 people opted for mud run and 'B' be the event that 60 people opted for river rafting. 30 people opted for both.
(End problem)
The problem is about applying the definition of independent events, and about the related concept of conditional probability. Events A and B are independent if and only if
P(A)P(B) = P(A ∩ B)
P(A ∩ B) is the joint probability, the probability that both events happen. Events A and B are subsets of the sample space (set of possible outcomes), and their intersection A ∩ B is the set of outcomes where both A and B occur. A is the set of all outcomes in the sample space which have the property "A occurred".
This garbled question seems to provide P(A), P(B), and P(A ∩ B), but it uses the word "Event" in a way that makes little sense.
If A is "the event that 70% of the children like chocolate cupcakes", then each outcome in the sample space must specify the cupcake preferences of every child, and A is the set of all outcomes where 70% of children like chocolate cupcakes. That describes a very complicated outcome with no justification for such complexity. Also, we are not given P(A) at all.
So let's say an outcome is the result of determining one child's cupcakes preferences, event A is the set of all outcomes where a child likes chocolate cupcakes, P(A) = 70%, and event B likewise for lemon cupcakes with P(B) = 25%.
The joint probability is supposed to be 30%. That can't be, because liking both implies liking lemon, but only 25% like lemon.
So let's suppose the joint probability was intended to be 25% and the lemon probability 30%. Then P(A)P(B) = 0.7×0.3 = 0.21, less than the joint probability. The events are not independent.
Is P(A ∩ B) > P(A)P(B) reasonable? Yes. It reflects the case where both are pretty unlikely, but they tend to occur together. What about P(A ∩ B) < P(A)P(B)? Yes it also is reasonable, and reflects the case where both are fairly likely, say 45%, but the intersection is small, less than 20%.