Bunuel
John has a 0.8 probability of eating breakfast with bacon and eggs, and Mary has a probability p of doing yoga in the morning.
If these events are independent, select for
At least one the probability that at least one of these events occurs, and select for
p the probability of Mary doing yoga in the morning that would be jointly consistent with the given information. Make only two selections, one in each column.
Deconstructing the Question We are given two independent events:
1. John eats breakfast: \(P(J) = 0.8\).
2. Mary does yoga: \(P(M) = p\).
We need to select two values from the list \(\{0.2, 0.4, 0.5, 0.85, 0.9, 0.95\}\):
* One for \(p\).
* One for \(P(\text{At least one})\).
Step 1: Set up the Formula Theory: For independent events A and B, P(A or B) = P(A) + P(B) - P(A)P(B). Alternatively, using the complement rule:
\(P(\text{At least one}) = 1 - P(\text{None})\)
\(P(\text{None}) = P(\text{Not J}) \times P(\text{Not M})\)
Given \(P(J) = 0.8\), then \(P(\text{Not J}) = 1 - 0.8 = 0.2\).
Given \(P(M) = p\), then \(P(\text{Not M}) = 1 - p\).
So, the equation is:
\(P(\text{At least one}) = 1 - [0.2 \times (1 - p)]\)
\(P(\text{At least one}) = 1 - (0.2 - 0.2p)\)
\(P(\text{At least one}) = 0.8 + 0.2p\)
Step 2: Test the Options We need to find a value \(p\) from the list such that the result \((0.8 + 0.2p)\) is
also in the list.
Try \(p = 0.2\): Result = \(0.8 + 0.04 = 0.84\) (Not in list).
Try \(p = 0.4\): Result = \(0.8 + 0.08 = 0.88\) (Not in list).
Try \(p = 0.5\): Result = \(0.8 + 0.10 = 0.90\).
Is 0.9 in the list?
YES.
Let's check the others just in case:
Try \(p = 0.85\): Result = 0.97 (No).
Try \(p = 0.9\): Result = 0.98 (No).
Conclusion The only consistent pair is:
p = 0.5 At least one = 0.9 Answer Selection: Column \(p\):
0.5 Column \(\text{At least one}\):
0.9