Probablity

Question

Given a coin which gives HEADS with probability 1/4 and TAILS with 3/4. The coin is tossed k times. What is the probability that we get at least k/2 HEADS is less than or equal to?
(A) (1/2) ^k/5
(B) (1/2) ^k/2
(C) (1/3) ^k/2
(D) (1/5) ^k/2

                                          
(Explain how to solve the summation of binomial equation that comes up )

Comments

Ans is given as A)
See this https://www.geeksforgeeks.org/aptitude-probability-question-1/
I am not able to get the solution they have given.

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i also understood only the first two lines in the given Solution.

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P(head) = ¼
P(tails) = ¾

We can verify it by putting k = 4 ,
P(atleast 2 heads ) = 1 – (P(1 head ) + P(0 head)
                                = 1 – ( 4C1*(1/4)*(3/4)^3 + (3/4)^4 )
                                = 0.26 ( approx )
 

Only option A satisfies .

Answer 1

Let's solve for k=2,

Now the probability of getting atleast 1 (k/2=1) head = 3/4*1/4 + 1/4*3/4 + 1/4*1/4 =  .437

Now option b) .5, c) .33, d) .2 are wrong.

Hence option A is correct.

Comments

Even option A will not get you at 0.4 with k=2.

$(1/2)^{2/5}$ = 0.757.

Answer 2

http://www.cse.iitk.ac.in/users/sbaswana/CS648/Lecture-2-CS648.pptx 

refer this ppt  slide number 4