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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8

1 2 3 4

Veerenk

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15 Dec 2025, 18:45

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Event-A: John eating BF with bacon & eggs, Probability, P(A): 0.8
Event-B: probability of Mary doing yoga in the morning, P(B): p

The probability of at least one of these events: Sum of the following events -
Probability of only Event -A & No event-B: P(A) X (1-P(B)): 0.8 (1-p)
Probability of only Event-b & No event -A: P(B) X (1-P(A)): p (1-0.8): 0.2p
Probability of both events: P(A) X P(B): 0.8p

Probability of At least one event: 0.8(1-p) + 0.2p + 0.8p= 0.8 + 0.2p

If p=0.2, P(At least one event): 0.8 + 0.04= 0.84 ==> Not consistent
If p=0.4, p(At least one event): 0.8 + 0.08= 0.88 ==> Not consistent
If p=0.5, P(At least one event): 0.8 + 0.1= 0.9 ==> Consistent
If p=0.85, P(at least one event): 0.8+ 0.17= 0.97 ==> Not consistent
If p=0.9, P(at least one event): 0.8 + 0.18= 0.98 ==> Not consistent
If p=0.95, P(at least one event): 0.8 + 0.19: 0.99 ==> Not consistent

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.

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15 Dec 2025, 19:17

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Bunuel

For independent events, the probability that at least one occurs follows the formula:

P ( at least one ) = 1 − P ( neither ) = 1 − ( 1 − 0.8 ) ( 1 − p ) = 0.8 + 0.2 p

Testing each value of p from the given options against this formula shows that when p = 0.5,
we get P(at least one) = 0.8 + 0.2(0.5) = 0.9,
and both values appear in the answer choices.

Selections At least one: 0.9 p: 0.5 These are the only two values from the provided list that are jointly consistent with the independent probability relationship.

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15 Dec 2025, 20:04

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John: 0.8
Mary: p

"At least one" = 1 - None = 1 - (1-0.8)(1-p) = 1 - 0.2 + 0.2p = 0.8 + 0.2p

Only matches with p=0.5 and "At least one"=0.9

"At least one"=0.9
p=0.5

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15 Dec 2025, 20:22

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Using set theory, probability at least one = p(A) + p(B) - p(A intersect B) = 0.8 + p - 0.8p = 0.2p + 0.8

Sustituting p with the values there is only one solution for p=0.5 and At least one=0.9

At least one=0.9 and p=0.5

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15 Dec 2025, 20:43

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Because the events are independent

The probability of both events happening at the same time is p(A&B) is = p(A)*p(B)

Therefore from the options
0.4 and 0.5 are to be chosen which will satisfy

Bunuel

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15 Dec 2025, 20:47

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probability John=0.8 -> probability no John=0.2
probability Mary=p -> probability no Mary=1-p

probability John or Mary = 1 - probability (no John and no Mary) = 1 - 0.2(1-p) = 0.8 + 0.2p

Using the given values, p=0.5 and at least one=0.9

At least one=0.9
p=0.5

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15 Dec 2025, 21:10

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p(John)=0.8 and p(Mary)=p

As the events are independent, p(John and Mary)=0.8p

p(at least one) = p(John or Mary) = p(John) + p(Mary) - p(John and Mary) = 0.8 + p - 0.8p = 0.2p + 0.8

For p=0.5 "At least one"=0.9. The other values doesn't work.

"At least one"=0.9 and p=0.5

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Bunuel

John: Probability of eating breakfast with egg & bacon (8/10 = 0.8); Probability of not eating breakfast with egg and bacon (2/10 = 0.2)
Mary: Probability of doing yoga p (1/2 = 0.5), Probability of not doing yoga (1/2 = 0.5)

1) At least one occurs: Both occur (0.8+0.5 = 1.3); One occurs (0.8-0.5=0.3); None occurs (0.2+0.5=0.7); Final: 1.3+0.3-0.7 = 0.9
2) Probability of Mary doing yoga = 1/2 or 0.5

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15 Dec 2025, 22:26

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What's the probability of at least one event happening? Well, it's the opposite of the probability of none of them happening.

Therefore:

Probability of John not eating breakfast \(= 1 - 0.8 = 0.2\)

Probability of Mary not going for yoga \(= 1 - p\)

And the joint probability of them not doing either = \(0.2 * (1-p)\)

Hence, the probability of at least one occuring is \(X = 1 - 0.2(1-p) = 1 - 0.2 + 0.2p = 0.8 + 0.2p\)

From here, we can insert potential values for p:

p = 0.2 =>\( X = 0.8 + 0.04 = 0.84\)

p = 0.4 => \(X = 0.8 + 0.08 = 0.88\)

p = 0.5 => \(X = 0.8 + 0.1 = 0.9\) which fits!

Therefore, the answers are: \(p=0.5\) and join probability \(X = 0.9\)

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15 Dec 2025, 23:15

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Probability of John having bacon & eggs for breakfast = 0.8

Probability of him not having = 1-0.8=0.2

Probability of at least one of them having eggs & bacon = 1- Probability of none of them having eggs & bacon

1-0.2*(1-p) (Assuming Mary has a probability of P)

Analyzing the given options we see that if we put p=0.5 we get (1-p=0.5) & the value 1-0.2*0.5 = 1-0.1 = 0.9 & both of them are in the given options

Hence the values are

At least one of them = 0.9
Mary's probability= 0.5

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P(B&E)=0.8 so, not P(B&E)=1-0,8=0,2
P(Y)=p so not P(Y)=1-p
Here, B&E -> Bacon and Eggs event, Y----> doing yogo event.

At least "one occurs" means we should subtract "both do not occur" : 1-not P(B&E)*not P(Y)=1-0,2*(1-p)=0.8+0,2p--> at least one happens.
0.8+0.2*p>=0.8 so, if it is 0.9, then 0.9=0.8+0.2p, p=0.5. So (0.9, 0.5)

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The probabity of doing atleast one event = 1- probability of none of events
=1-(1-0.8)(1-p)
=1-0.2*(1-p)
=0.8+0.2p > 0.8
substitute the values: 0.85
0.05=0.2p
p=0.25
No Option

Substitute the value as 0.9
0.8+0.2p=0.9
0.1=0.2p
p=0.5.
Satisfy the given values.

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16 Dec 2025, 00:34

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Probability of eating breakfast with bacon and eggs = 0.8
Probability of not eating breakfast with bacon and eggs = 1 - 0.8 = 0.2
Probability of Mary doing yoga in the morning = p
Let probability of Mary not doing yoga in the morning = p`= 1 - p
Probability of atleast one of these events occurring, P1 = 1 - probability of none of these events occurring
= 1 - (0.2)*p`

Using trial and error,
a) let p = 0.2, then P1 = 1 - (0.2)*(0.8) = 1 - 0.16 = 0.84 - NOT present in options
b) let p = 0.4, then P1 = 1 - (0.2)*(0.6) = 1 - 0.12 = 0.88 - NOT present in options
c) let p = 0.5, then P1 = 1 - (0.2)*(0.5) = 1 - 0.1 = 0.9 - available in options
So p = 0.5, atelast one = 0.9

Bunuel

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A John b&e= P (A)=0.8
B Mary yoga P (B) =p

P (at least one) = 1 - P (neither)
= 1 - (1-P(A)) (1-P(B))
=1-(0.2) (1-P)
1-0.2+0.2P=0.8+0.2P

IF P=0.2, 0.8+0.2(0.2) =0.84
P=0.4=0.4, 0.8+0.2(0.4) =0.88
P=0.5, 0.8+0.2(0.5) =0.9
P=0.85, 0.8+0.2(0.85) =0.97
P=0.9, 0.8+0.2(0.9) =0.98
P=0.95, 0.8+0.2(0.95) =0.99

SO At least one= 0.9
p=0.5

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I was stuck on this one for a while (explains my timing issues with DI). Anyway, the key is to quickly realize that if we're looking at the probability for either one or both of the outcomes occurring, we can easily find the probability of neither outcome occurring and subtract that from 1, as the remainder will be the required probability.

John has a 0.8 probability of eating breakfast, or a 0.2 probability of not. We don't have Mary's probability but we can look to the answer choices to get a sense. I'd go from smallest to largest. If 0.2 is her probability of yoga, 0.8 is her probability of not. 0.8*0.2 = 0.16, but 1 - 0.16 or 0.84 isn't there in the results.

Then I move to 0.4, or 0.6 of not doing yoga. 0.2*0.6 = 0.12. 1 - 0.12 or 0.88 is not among the answers either.

0.5 - or 0.5 of not doing yoga. 0.2*0.5 = 0.1. 1 - 0.1 = 0.9 is there is in the choices. We can park this and check the remaining ones, to be sure.

0.85 - or 0.15 of not doing yoga. 0.2*0.15 = 0.03. 0.97 is not among the answers.
0.9 - or 0.1 of not doing yoga. 0.2*0.1 = 0.02. 0.98 is not among the answers.
0.95 or 0.05 of not doing yoga. 0.2*0.05 = 0.01. 0.99 is not among the answers.

We can lock our answer in then: At least one will be 0.9, and p will be 0.5.

Bunuel

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16 Dec 2025, 01:23

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John's Failure probability = 1 - 0.8 = 0.2
Mary's Failure probability is P(M ) = 1 - p
P (Both Fail = 0.2(1-p)
Atleast one = 1- (0.2(1-p))

Now let's walk through options
Option p = 0.5
Mary fails 50% of the time
50% of 0.2 remains
P = 1- 0.1 = 0.9 ✅

Option p = 0.4
60% of 0.2 remains
1-0.12 = 0.88 not an option

Similarly try for other options .

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16 Dec 2025, 02:44

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Given-
P(q)=0.8 & p is the probability of doing yoga.
To find=
1) Probability of at least one event
2) Probability of doing yoga, (p)

When events are independent,
=> P (at least one)=P(q)+P(p)-P(p)P(q)
=> 0.8+p-0.8p=0.8+0.2P

Checking the answer choices,
we get p=0.5, which gives P(at least 1)=0.8+0.2(0.5)=0.9

Therefore, 0.9 & 0.5 are the correct choices.

Bunuel

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16 Dec 2025, 03:00

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Bunuel

John eating breakfast with bacon and eggs = 0.8
Mary doing yoga = p

John not eating eggs and bacon with breakfast = 1 - 0.8 = 0.2
Mary not doing yoga = 1 - p

Doing at least one = 1 - Not doing both = 1 - (0.2)(1-p)
=> 1 - 0.2 + 0.2p

We need to find the value of p such that the above equation can be satisfied

if p = 0.8 we get At least one = 1 - 0.2 - 0.2*0.8 = 1 - 0.2 + 0.16 = 1- 0.04 = 0.96

Similarly doing for all we get p = 0.5 gives us At least one = 1 - 0.2 + 0.2*0.5 = 0.8 + 0.1 = 0.9

This satisfies by
p = 0.5 and
At least one = 0.9

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For independent events, P(A and B) = P(A) * P(B)
P(At least one) = P(John breakfast) + P(Mary yoga) - P(John breakfast) * P(Mary yoga) = 0.8 + p - 0.8p = 0.8+0.2p
0.9 = 0.8 + 0.2*0.5
P(At least one) = 0.9
p = 0.5