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

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

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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 \(= (\frac{1}{2)}^5*[5C_2+5C_1+5C_0] = 16*(\frac{1}{2})^5 = 1/2\)

Answer: Option E

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

­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