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

1. $(.4)^{10}$
2. $1 - (.4)^{10}$
3. $1 - (.6)^{10}$
4. $(.6)^{10}$
5. $10(.4) (.6)^{9}$

$10$ 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: $C$

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

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...

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....

nice explanation digvijay sir
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 – $(0.6)^{10}$