A certain event has only outcomes A or B, and they are equally likely

Question

A certain event has only outcomes A or B, and they are equally likely and mutually exclusive. Out of 5 random occurrences of the event, what is the probability that fewer than half result in an outcome of A?

A. 3/16

B. 5/16

C. 1/3

D. 2/5

E. 1/2

A. 3/16
B. 5/16
C. 1/3
D. 2/5
E. 1/2

Kinshook · Answer

Probability (0A & 5B) = (1/2)^5 = 1/32
Probability (1A & 4B) = 5C1 * (1/2)(1/2)^4 = 5/2^5 = 5/32
Probability (2A & 3B) = 5C2 * (1/2)^2(1/2)^3 = 10/32 
Probability (0A or 1A or 2A) = 16/32 = 1/2

IMO E

satya2029 · Answer

Total outcome=32
for 0-A outcome= !5/!5=1
for1-A outcome=!5/!1*!4=5
for 2-A outcome=!5/!2*!3=10
Required Probability=(1+5+10)/32
=1/2
E:)

gurmukh · Answer

Everything is symmetrical so answer is straight away 1/2.
No calculation required.

Posted from my mobile device

GMATinsight · Answer

Probability of A to happen = 1/2
Probability of B to happen = 1/2

Total occurrences = 5

Fewer than half = 2 or 1 or 0

Probability of 2 a and 3 B  = 5C_2*(\frac{1}{2})^2*(\frac{1}{2})^3

Probability of 1 A and 4 B  = 5C_1*(\frac{1}{2})^1*(\frac{1}{2})^4

Probability of 0 A and 5 B  = 5C_0*(\frac{1}{2})^0*(\frac{1}{2})^5

Total Probability  = (\frac{1}{2)}^5*  = 16*(\frac{1}{2})^5 = 1/2

Answer: Option E

Fdambro294 · Answer

From a logical standpoint, because there are 5 balls and the balls must be integers, there are 2 complementary events:

Either:

(More A balls than B balls) or (More B balls than A balls)

For example, for any outcome in which there are more A balls than B balls

-greater than (1/2) of the balls will always be A balls

And

-less than (1/2) of the balls will always be B balls

Out of every possible outcome, symmetrically:

(1/2) of the outcomes will have more A balls (A balls will be greater than > (1/2) of the balls in the outcome)

And

(1/2) of the outcomes will have more B balls (A balls will be less than < (1/3) of the balls in the outcome)

Probability = (1/2)

That on any outcome, the A balls will be fewer than (1/2) of all the balls

Posted from my mobile device

ShuvoDhk · Answer

Hi Kinshook, thanks for your sugegstions. But I'm stuck in below situation. Can you please correct me with clarification.
Desired Events = 0A & 5B OR 1A & 4B OR 2A & 3B
                 =(5C0/32 * 5c5/32) + (5c1/32 * 5c4/32) + (5c2/32 * 5c3/32) {as everytime total possible outcome remains same i.e. 32}
                 = (1/32 * 1/32) + (5/32 * 5/32) = (10/32 * 10/32)
                 = 1/1024 + 25/1024 + 100/1024 = 126/1024

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A certain event has only outcomes A or B, and they are equally likely

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