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 )
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.
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.
Even option A will not get you at 0.4 with k=2.
$(1/2)^{2/5}$ = 0.757.
http://www.cse.iitk.ac.in/users/sbaswana/CS648/Lecture-2-CS648.pptx
refer this ppt slide number 4
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