TIFR CSE 2010 | Part A | Question: 6

Question

Given 10 tosses of a coin with probability of head = .$4$ = ($1$ - the probability of tail), the probability of at least one head is?

• A. $(.4)^{10}$
• B. $1 - (.4)^{10}$
• C. $1 - (.6)^{10}$
• D. $(.6)^{10}$
• E. $10(.4) (.6)^{9}$

Answer 1

$10$ tosses of coin are there.

Probability of head = $0.4$

Probability of tail = $0.6$

Probability of at least one head $=1- P{_\text{no head occur}} ={1-(0.6)^{10}}.$

Correct Answer: $C$

Comments

wat is the problem with option E ?? can u explain wat u hav done ?? how is it (0.6)¹⁰ ??

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@ 

first thing..    X ( wat)->what  :)

and second 

10(.4)(.6)^{9   .}.it is 10C1 (.4)(.6)⁹  ...and according to this.... one toss out of 10 should be head....

but the question is for at least one...here head could be more than one...

correct answer...

1-(.6)¹⁰

here we have find the probability of 0 head....menas all 10 toss must  be tail....

and negation of it will return at least one head....

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nice explanation digvijay sir

Answer 2

n = 10

Using Binomial expansion,

probability of getting success(head) = 0.4

probability of getting failure(tail) = 1-0.4 = 0.6

probability of getting at least one head P(x >=1) = 1 – P(x=0) = 1 – $\binom{10}{0}(0.4)^{0}(0.6)^{10-0}$ = 1 – $(0.6)^{10}$

Answer C