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655-705 (Hard)|   Math Related|         
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 03:21
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A: Probability of JOhn eating breakfast = 0.8
B: Prob of Mary doing yoga = p

let y = P(atleast 1) = P(A or B) = P(A) + P(B) - P(A and B)
P(A and B) = 0.8p as A and B are independent events
So y = 0.8 + p - 0.8p
y = 0.8+ 0.2p
p=0.5 and y = 0.9 from the options satisfy this.
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 04:00
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Equation to solve : 0.8 + p - 0.8*p

Use, p = 0.5 ; then at least one = 0.9
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 05:14
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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.

Probability of John eating breakfast= 0.8
Probability of Mary doing yoga = p
The events are given to be independent
Also, P(atleast 1) = 1-P(neither)
P(neither) = (1-0.8)(1-p)= 0.2(1-p)
P(atleast 1)= 1-0.2(1-p) = 0.8+0.2p
Try to put the values of p given in choices
p=0.2 ; P(atleast1)= 0.84 ....not in options
p=0.4 ; P(atleast1)= 0.88.....not on option
p=0.5 ; P(atleast 1)=0.90 .....yes

P(atleast 1)=0.90
p=0.5
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 06:11
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Probab (For at least 1)= 1- P(none)

Check,

p=0.2, P(at least 1)= 1- (1-0.8)(1-0.2)=.84 no option
p=0.4, P (at least 1)= 1- (1-0.8)(1-0.4)= .88 no option
p=0.5, P (at least 1)= 1- (1-0.8)(1-0.5)= 0.9 Matched

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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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Rahilgaur
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 06:34
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Bunuel
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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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Probability that At least of the these event occurs = 1- {probability None of the event occurs} = 1 - {0.2*(1-p)} = x (Let's assume)
Putting available answers of P to find the consistent answer with above

If p = 0.2 ---> x= 1- 0.2*0.8= 1-0.16 = 0.84
p=0.4 ----> x = 1- 0.12 = 0.88
p= 0.5 -----> x= 1-0.1 = 0.9 (Answer)
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 06:38
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For atleast 1 questions answer is mostly 1- p(event never occurs)

so P(John eating breakfast) = 0.8 P(John not eating breakfast )=0.2
similarly P(Mary doing yog)=p P(mary not doing yoga)=1-p
Since both events ar eindependent P(John not breakfast and Mary not yoga) =0.2(1-p)
As mentioned in the begining P(atleast 1)=1-0.2(1-p) =>0.8+0.2p=>0.2(4+p)
only option which works is p=0.5 and P(atleast 1)=0.9
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 06:39
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Bunuel
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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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Both the events are indepent.
Also if we want the prob of atleast 1 then we can say that,
prob(at least one) = 1 - prob(neither)

Therefore, prob(at least one) = 1 - ((1-0.8)(1-p))
= 1 - 0.2*(1-p)
= 0.8 + 0.2p

Now, use options to find out the correct consistent pair.

After putting p = 0.5
We get prob(at least one) = 0.9

Therefore,
at least one = 0.9
and p = 0.5
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 07:05
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Bunuel
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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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John - Yes - 0.8
John - No - 0.2

MAry - Yes - p
Mary - No - (1 - p)

We need

(John Yes and Mary No) or (John No and Mary Yes) or (John Yes and Mary Yes)
[0.8 (1-p)] + 0.2p + 0.8p
--> 0.8 + 0.2p

Choose p from the options and the corresponding value for the overall probability.

Out of the options, p = 0.2 gives us overall probability = 0.84

Only one other option p = 0.95 gives overall probability = 0.89.

Both appear to be close. I choose 0.2 for p and 0.84 overall probability.
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 07:19
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Probability that John eats breakfast = 0.8, say (A)
Probability that Mary does Yoga = p, say (B)
Probability that one of these occur is P(A U B), i.e. P(A) + P(B) - P(A n B)

This gives, 0.8 + p - 0.2p = 0.8 + 0.2p
As per the options given, only one fits, i.e. 0.8 + 0.2 x (0.50) = 0.90
Hence, at least one of the events occurring is 0.90

As for p, no value has been given and it is an independent event, hence probability of occurring will be 1/2, i.e. 0.50.
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 07:47
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We are given,
P(J) = Probability Jogn eats bacon and eggs = 0.8
P(M) = Probability Mary does yoga = p
Both these events are independent

Now we see that,
=> P(J') = P(John doesent eat) = 1 - 0.8 = 0.2
=> P(M') = P(Mary doesent do yoga) = 1 - p

Given the events are independent, P(neither) = 0.2*(1-p)

Since we know that,
P(atleast one) = 1 - P(neither)
=> P(atleast one) = 1 - 0.2*(1-p) = 0.8 + 0.2p

We want to find out P(atleast one), and from the given options of P(atleast one) and p,
the values which match with the given 0.8 + 0.2p = P(atleast one) is
p = 0.5
P(atleast one) = 0.9 (0.8 + 0.2*0.5 - 0.9)
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 08:01
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John probability 0.8
Mary probability p
Events are independent

P(at least one) = 1- P(neither)
P(at least one ) = 1- 0.2(1-p)

Now, while checking the values for p and At least one that jointly consistent are 0.5 & 0.9
p is 0.5
at least one is 0.9
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 08:04
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Given =>

Probability of having bacon with eggs => 0.8
Probability of doing yoga => p

Probability of at least 1 happening => 1 - probability of none happening --- 1
As these are independent events, probability of both not happening = (1-0.8)*(1-p) => 0.2-0.2p ---2

=> Using 1 & 2

Probability of at least 1 happening => 1-0.2+0.2p => 0.8+0.2p

Only values satisfying => Atleast 1 - 0.9 ; p=>0.5
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12 Days of Christmas GMAT Competition 2025 - Day 1: John has a 0.8 16 Dec 2025, 08:46
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Prob(a) of John having bacon and eggs = 0.8
prob(b) of mary doing yoga = p
Prob at least on of them occurs = P(A) + P(B)- P(A)P(B)
0.8 + p - (0.8)p = 0.8 +0.2p
after plugging in values
p=0.5
at least one = 0.9
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