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

  1. 5/32
  2. 15/128
  3. 35/128
  4. None of the options
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For $7^{th}$ toss head appears exactly $4$ times means in the previous $6$  tosses head can appear at any position but exactly thrice and the $7^{th}$ toss will result in a head.

Required probability  $= \binom{6}{3}\left ( \frac{1}{2} \right )^{3}\left ( \frac{1}{2} \right )^{3}\left ( \frac{1}{2} \right )$  $= \frac{5}{32}$             

$A$ is correct.

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→ 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=½ and probability of a tail=½
→ 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
6C3*(½)3 *(½)3= 5/16 (3 success in 6 trials can happen in 6C3 ways). → 7th toss, The probability of obtaining a head=1/2
→ In given question is not mention that what happens after 7 tosses.
→ Probability for that for 7th toss head appears exactly 4 times is =(5/16)*(1/2)
=5/32.

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