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If a random coin is tossed $11$ times, then what is the probability that for $7$th toss head appears exactly $4$ times?

1. $5/32$
2. $15/128$
3. $35/128$
4. None of the options

To find probability that for the 7th toss head appears exactly 4 times. We have find that past 6 run of tosses.
Past 6 tosses, What happen we don't know. But according to 7th toss, we will find that head appeared for 3 times.
Any coin probability of a head=$1/2$ and probability of a tail= $1/2$

If we treat tossing of head as success, then this leads to a case of binomial distribution.
According to binomial distribution, the probability that in 6 trials we get 3 success is:

$\binom{6}{3}*(1/2)^{3} *(1/2)^{3}= 5/16$

7th toss, The probability of obtaining a head=$1/2$

In given question is not mention that what happens after 7 tosses.
so Probability for that for 7th toss head appears exactly 4 times is

=$(5/16)*(1/2)$
=$5/32$
Here we have been asked to find the probability of getting exactly a head at the 7th toss and 3 heads in any other 6 tosses. So all over 4 heads in the 7 tosses.

Please understand, it is different from the question  “What is the probability that for 7 tosses, the head appears exactly 4 times?” In that case head would have been appeared in any 4 out of 7 and the prob would have been $\binom{6}{4}$ * $\frac{1}{2^{4}}$ * $\frac{1}{2^{3}}$ = $\frac{35}{128}$

But here in this problem, first we have to find the prob of getting 3 heads out of 6 tosses. and this will be

= $\binom{6}{3}$ * $\frac{1}{2^{3}}$ * $\frac{1}{2^{3}}$

= $\frac{5}{16}$

and for the 7th toss, prob of head is $\frac{1}{2}$

Hence the prob becomes  $\frac{5}{16} * \frac{1}{2}$ = $\frac{5}{32}$

Hence, ans is option A.
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