# NIELIT DEC 2017 SET-C 62

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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?

A) 5/32

B) 15/128

C) 35/128

D) None of the options

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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What about the tosses done from 8,9,10,11 do we need to ignore them ? if to ignore why so ?
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yes it is redundant information, here we're more concerned about the 7th toss, it might even be possible that coin is tossed infinite times
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Thanks got it .
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Official answer is C for this question.
1 vote
→ 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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