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15 Dec 2025, 10:56
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This wouldn't take more than 30 seconds if you just get the basic idea in place.
For at least one, let's take 1 - (both events did not occur)
Probability that John didn't eat bacon and eggs for breakfast = 0.2 (1-0.8)
Look at the options, try with different values.
If p = 0.2
p not happening = 0.8
Probability that both events don't happen = 0.2*0.8 = 0.16.
Probability of at least one happening =0.84 (which is not in the option). Ignore p = 0.2.
Let's check p = 0.5
p not happening = 0.5
Probability that both events don't happen = 0.2*0.5 = 0.1
Probability of at least one happening = 0.9
Therefore, at least one is 0.9, and p is 0.5
For at least one, let's take 1 - (both events did not occur)
Probability that John didn't eat bacon and eggs for breakfast = 0.2 (1-0.8)
Look at the options, try with different values.
If p = 0.2
p not happening = 0.8
Probability that both events don't happen = 0.2*0.8 = 0.16.
Probability of at least one happening =0.84 (which is not in the option). Ignore p = 0.2.
Let's check p = 0.5
p not happening = 0.5
Probability that both events don't happen = 0.2*0.5 = 0.1
Probability of at least one happening = 0.9
Therefore, at least one is 0.9, and p is 0.5
15 Dec 2025, 11:06
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Took me a while....but I guess.. the best is to make a quick column and quickly solve.........
| If p(Yoga) | then (1-p)not yoga | Breakfast 0.8 | not breakfast 0.2 | Atleast 1..is 1 - both not happening | |
| 0.2 | then 0.8 | 0.8 | 0.2 | 1 - (0.8 x 0.2) - Answer will have 4 in the end..not in option | |
| 0.4 | then 0.6 | 0.8 | 0.2 | 1 - (0.6 x 0.2) - Answer will have .8 in the end ..not in option | |
| 0.5 | 0.5 | 0.8 | 0.2 | 1 - (0.5 x 0.2 ) = 0.9 (Answer but lets check others) | |
| 0.85 | 0.15 | 0.8 | 0.2 | 1 - (0.15 x 0.2 ) = 1 - 0.03 = 0.97 (not in Option ) | |
| 0.9 | 0.1 | 0.8 | 0.2 | 1 - (0.1 x 0.2) = 0.98 (The probability of at least 1 happening will only increase from here.as the value of p increases and ther is no option greater than 0.95 ..so use of doing it.. |
Bunuel
15 Dec 2025, 11:09
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Let probability of john's event happening be p(J) and probability of Mary's event happening be p(M)
firstly, since J and M are independent events:
\(p(J\text{ and }M) = p(J) * p(M)\)
What is the significance of calling them out as independent? Because, if J and M are dependent events, then:
\(p(J\text{ and }M) = p(J) * p(M \text{ | } J)\)
for at least one happening:
\(p(J\text{ or }M) = p(J) + p(M) - p(J \text{ and } M)\)
since J and M are independent events:
\(p(J \text{ or } M)= p(J) + p(M) - p(J) * p(M)\)
given that \(p(J) = 0.8\) and \(p(M) = p\), so substituting those
\(= 0.8 + p - (0.8 * p)\)
\(= 0.8 + 0.2p\)
substitute different values for \(p\) from the answers
if \(p = 0.2\), then
\(p(J \text{ or } M) = 0.8 + 0.2 * 0.2\)
\(= 0.8 + 0.04\)
\(= 0.84 \)
But this is not in the list of answers for At least one
if \(p=0.4\), then
\(p(J \text { or } M) = 0.8 + 0.2 * 0.4\)
\(= 0.8 + 0.08\)
\(= 0.88\)
But this is not in the list of answers for At least one
if \(p = 0.5\), then
\(p(J \text{ or } M) = 0.8 + 0.2 * 0.5\)
\(= 0.8 + 0.1\)
\(= 0.9\)
This exists for At least one.
So ans: \(0.9, 0.5\) or \(E, C\)
firstly, since J and M are independent events:
\(p(J\text{ and }M) = p(J) * p(M)\)
What is the significance of calling them out as independent? Because, if J and M are dependent events, then:
\(p(J\text{ and }M) = p(J) * p(M \text{ | } J)\)
for at least one happening:
\(p(J\text{ or }M) = p(J) + p(M) - p(J \text{ and } M)\)
since J and M are independent events:
\(p(J \text{ or } M)= p(J) + p(M) - p(J) * p(M)\)
given that \(p(J) = 0.8\) and \(p(M) = p\), so substituting those
\(= 0.8 + p - (0.8 * p)\)
\(= 0.8 + 0.2p\)
substitute different values for \(p\) from the answers
if \(p = 0.2\), then
\(p(J \text{ or } M) = 0.8 + 0.2 * 0.2\)
\(= 0.8 + 0.04\)
\(= 0.84 \)
But this is not in the list of answers for At least one
if \(p=0.4\), then
\(p(J \text { or } M) = 0.8 + 0.2 * 0.4\)
\(= 0.8 + 0.08\)
\(= 0.88\)
But this is not in the list of answers for At least one
if \(p = 0.5\), then
\(p(J \text{ or } M) = 0.8 + 0.2 * 0.5\)
\(= 0.8 + 0.1\)
\(= 0.9\)
This exists for At least one.
So ans: \(0.9, 0.5\) or \(E, C\)
