12 Days of Christmas GMAT Competition - Day 10: Elvin has sent wish

Elvin has sent wish lists to both Santa and Mrs. Claus. What is the probability that he will receive gifts from neither of them?

(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4. --> not enough
(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both. --> not enough

E

P of neither
P (s) is probability of gift from Santa and P(m) is gift from Mrs Santa

#1

The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.


P (s)+P(m)-P (both) =1
P(s)+P(m)=5/4
neither P will be 1-5/4 ;
1/4
sufficient
#2
The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.
P(s)= 3*P(both)
we get P(s) = 1/2 and P (m) as 1/2
and not getting gift is also 1/2
1/2 * 1/2 ; 1/4
sufficient
[color=#00aeef]option D[/color]

Elvin has sent wish lists to both Santa and Mrs. Claus. What is the probability that he will receive gifts from neither of them?
(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.
(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.

Sol: IMO Option-C

P(Exactly One) + P(Both) + P(Neither) = 1 -----(1)

P(Neither) = ?

Stat-1:
The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.
P(Both) = 1/4
No other info. This statement is Insufficient.

Stat-2:
The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.
P(Exactly One) = 3*P(Both)
Using this in (1) = P (Neither) = 1-4*P(Both)
This depends on the P(Both) and may vary.
Therefore, This statement is Insufficient

(1) and (2)
Using P(Both) from stat-1 and using in stat-2 solves for P(Neither) = 0
Sufficient

To determine the probability that Elvin will receive gifts from neither Santa nor Mrs. Claus, we need to find the complement of the event that he receives gifts from at least one of them.

Let's define the events:
A: Receiving gifts from Santa
B: Receiving gifts from Mrs. Claus

We want to find P(AB), the probability that Elvin does not receive gifts from either Santa or Mrs. Claus.

Statement (1) tells us that P(A∩B) = 1/4, the probability that Elvin receives gifts from both Santa and Mrs. Claus. However, it does not provide any information about the individual probabilities P(A) and P(B). Without knowing the individual probabilities, we cannot determine P(AB). Statement (1) is insufficient.

Statement (2) states that P(A∪B) = 3P(A∩B), the probability that Elvin receives gifts from at least one of them is three times the probability that he receives gifts from both. This implies that P(A∩~B) = 2P(A∩B) and P(~A∩B) = 2P(A∩B). However, it does not provide information about the individual probabilities P(A) and P(B). Without knowing the individual probabilities, we cannot determine P(AB). Statement (2) is insufficient.

Considering both statements together, we have information about the relationship between the probabilities P(A), P(B), and P(A∩B). However, we still lack the specific values of P(A) and P(B) to determine P(AB). Therefore, both statements together are insufficient.

The answer is (E) - both statements together are insufficient to determine the probability that Elvin will receive gifts from neither Santa nor Mrs. Claus.

Given: Elvin has sent wish lists to both Santa and Mrs. Claus.
Asked: What is the probability that he will receive gifts from neither of them?

Let the probability of receiving gift from Santa be x and the probability of receiving gift from Mrs. Claus be y.

(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.
The probability that he will receive gifts from both Santa and Mrs. Claus = x*y = 1/4 = 1*1/4 = 1/4*1 = 1/2*1/2
The probability of receiving gifts from neither of them = (1-x)(1-y) = 0 or 1/4
NOT SUFFICIENT

(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.
x + y -xy = 3xy
x + y = 4xy ;
The probability that he will receive gifts from neither of them = (1-x)(1-y) = 1 - x - y + xy = 1 - 3xy
NOT SUFFICIENT

(1) + (2)
(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.
x*y = 1/4
(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.
x + y - xy = 3xy
x + y = 4xy
The probability that he will receive gifts from neither of them = (1-x)(1-y) = 1 - x - y + xy = 1 - 3xy = 1 - 3*1/4 = 1 - 3/4 = 1/4
SUFFICIENT

IMO C

Here,
A: the event that Elvin receives gifts from Santa.
B: the event that Elvin receives gifts from Mrs. Claus.

The probability of the union of two events is given by the inclusion-exclusion principle:

P(A U B) = P(A) + P(B) - P(A and B)

Now, the probability that Elvin receives gifts from neither of them is the complement of the union of events A and B :

P(neither A nor B) = 1 - P(A U B)

Let's analyze the given statements:

(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4

This gives us P(A and B) = 1/4 -> Not Sufficient

(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.

Let p be the probability that he receives gifts from both, then the probability that he receives gifts from exactly one is 3p .

Now, we can set up an equation using the inclusion-exclusion principle:

P(A U B) = P(A) + P(B) - P(A and B) =3p + p =4p -> Not Sufficient

Combining both,we get

P(A U B) = 4 * 3/4 = 1

which suggests that Elvin will certainly receives gift from one of them

So, C is correct answer IMO

1) This statement alone doesn't provide enough information to determine the probability that Elvin will receive gifts from neither of them.
We don't know the probabilities of the other outcomes (receiving a gift from only Santa, only Mrs. Claus, or neither).
2) This statement also doesn't provide enough information on its own. We still don't know the probability of receiving gifts from neither.

1+2
Let's denote the probability of receiving gifts from both as B, from exactly one as O, and from neither as N.
From statement (1), we know that B = 1/4.
From statement (2), we know that O = 3B = 3/4.
Since the sum of all probabilities must be 1, we have B + O + N = 1.
Substituting: 1/4 + 3/4 + N = 1
N = 1 - 1 = 0.
So the probability that Elvin will receive gifts from neither Santa nor Mrs. Claus is 0.

Hence C

A+B+C+D=1
What is D?

S1) C=1/4 (Not suff)

S2) A+B=3C (not Suff)

S1+S2) C=1/4, A+B=3/4 gives D=0

(Suff)

C)

(C) imo
(1)inconclusive alone
(2)inconclusive
Combine both yes we know individual chance is 3/4 where as from both is 1/4 from neither thus is zero so (C).

Assuming
P(A) = getting gift from only Santa
P(B) = getting gift from only claus
P(C) = getting gift from both
P(D) = not getting any gift.

P(A)+P(B)+P(C)+P(D) = 1

Statement 1

P(C) = 1/4

Not sufficient

Statement 2

P(A) + P(B) = 3P(C)

Not sufficient

Statement 1 + Statement 2

1+P(D) = 1
P(D) = 0

Sufficient

OA should be C.

Posted from my mobile device

Elvin has sent wish lists to both Santa and Mrs. Claus. What is the probability that he will receive gifts from neither of them?

Only Santa: a
Only Mrs Claus:b
Both: x
Neither: N
Total: T
T= a+b+x+N
Question: N/T?

(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.
x/T=1/4
x=k
T=4k
Question: N/T =N/4k?
-->Not sufficient

(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.

(a+b)/T=3*(x/T)
a+b=3x

Question: N/T?
-->Not sufficient

(1)+(2)
x=k
T=4k

a+b=3x
a+b=3k

T=a+b+x+N
4k=3k+k+N
N=0

Question: N/T=0/4k =0
--> Sufficient

Right answer: C

Elvin has sent wish lists to both Santa and Mrs. Claus. What is the probability that he will receive gifts from neither of them?

(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.
(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.

To find the probability that Elvin will receive gifts from neither of them, we need to find probability of receiving gift individually from Santa and Claus and also from both.

S1:- We can't find the individual probability in this case.
Insufficient.

S2:- We only get a relation between individual probability and from both.
Insufficient.

S1&S2:- We can find the individual probability of recieving gifts, and also receiving gifts from both.

C is correct choice.

x = probability that he will receive gifts from only Santa
y = probability that he will receive gifts from both
z = probability that he will receive gifts from only Mrs. Claus
n = probability that he will receive gifts from neither of them

We know that:

x+y+z+n=1 (a)

¿n?

(1) y=1/4

Clearly insufficient

INSUFFICIENT

(2) x+z=3y

Sustituting in (a):

n=1-4y

INSUFFICIENT

(1)+(2)

n=1-4y=1-1=0

The answer is 0

SUFFICIENT


IMO C

Maybe I am getting tripped up here, but I think the answer is C...

probability of neither would be 1 - P(A) - P(B)

(1) insufficient
This tells us P(AB) is 1/4 - so we know the probability of getting both, but nothing about them individually

(2) insufficient
This tells us P(A) OR P(B) (or P(A) + P(B) - P(AB) is 3 * P(AB) = 3*1/4 = 3/4

(1) and (2) sufficient

I think this is sufficient, as we know P(A) + P(B) must equal 1 because the probability of getting a present from either (A OR B) is 3/4 and probability of both is 1/4, so we have 1-1 = 0 so we know!

Am I missing something about calculating A/B?

Elvin has sent wish lists to both Santa and Mrs. Claus. What is the probability that he will receive gifts from neither of them?

(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4. Insufficient, one of them could have a probability of 1. we don't know
(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both. insufficient, we don't know either probability from this statement.
Sufficient together as we know the probability of both AND the probability of one of them, we can calculate NONE from this.

Answer: (E)

Setting A = probability of Santa gift, B = probability of Mrs. Claus gift, and C = probability of both, the probability that Elvin receives at least one gift is given by A+B-C. Then, the probability of receiving no gifts is given by 1-(A+B-C). We need information about these missing pieces to answer this question.

(i) Insufficient.
This tells us C = 1/4. This is not enough information as we don't know anything about A or B.

(ii) Insufficient.
This tells us that either A = 3C or B = 3C.
Substituting, we can get to: prob = 1-(3C+B-C) = 1-2C-B or prob = 1-(A+3C-C) = 1-A-2C.
Again, we still have 2 unknowns.

(i) + (ii) Insufficient.
Combining the two, we can get down to: prob = 1-2(1/4)-B = 1/2-B or 1-A-2(1/4) = 1/2-A.
Again, without information about A and/or B, still still cannot answer this question.

Elvin has sent wish lists to both Santa and Mrs. Claus. What is the probability that he will receive gifts from neither of them?
Let S and C represent the probability that Elvin will receive gift from Santa and Mrs Claus respectively.
S' and C' be the probability that Elvin will not receive gift from Santa and Mrs Claus respectively.
we need to find the probability of E'S' =?
(1) The probability that he will receive gifts from both Santa and Mrs. Claus is 1/4.
E*S = 1/4.
not sufficient as we don't know any other values.
(2) The probability that he will receive gifts from exactly one of them is three times the probability that he will receive gifts from both.
E*S' + S*E' = 3*E*S
we know that, E*S' + S*E' + E*S + E'*S' = 1 (as it is sum of probabilities)
=> 4*E*S +E'*S' = 1
We don't know the values of E*S
hence, not sufficient

both statements together
E*S = 1/4 and 4*E*S + E'*S' = 1
=> E'*S' = 1 - 4*(1/4) = 0
Hence, sufficient
Answer C