Question

A coin is tossed until one observes a sequence of exactly three tails. The probability that the experiment comes to an end at 7th toss is
A.$\dfrac{7}{{128}}$
B.$\dfrac{1}{{128}}$
C.$\dfrac{1}{{32}}$
D.$\dfrac{5}{{128}}$

Answer 1

Hint : For the experiment coming to an end at the 7th toss, one must get a tail at 5th, 6th and 7th toss. But in question, it is said that one must observe exactly three tails, therefore, at 4th toss, one must get a head. Also, a sequence of exactly three tails must not be observed before the 7th toss.

Complete step-by-step answer :
We know that Probability of an event E is given by $P(E) = \dfrac{{{\text{number of cases favourable to event E}}}}{{{\text{Total number of cases in sample space}}}}$
Since, for each tossed coin, we have 2 possible outcomes either a Head or a Tail.
Therefore, total number of cases in sample space is given by:
${\text{total number of cases}} = {2^7} \\\ = 128 \;$
In favorable cases, we must have a head at 4th toss and a tail at 5th, 6th and 7th toss i.e., _ _ _ H T T T.
For 1st, 2nd and 3rd toss, we have 2 possible outcomes in each toss, that is, either a head or a tail.
Therefore, total number of possible outcomes for these 3 tosses is given as:
${\text{total number of favourable cases}} = {2^3} \\\ = 8 \;$
But we will exclude the case in which the tail occurs in all the three tosses.
Therefore, the number of favorable outcomes is now$\left( {8 - 1} \right) = 7$.
Hence, Probability of experiment coming to an end at 7th toss is given by:
$P(E) = \dfrac{{{\text{number of cases favourable to event E}}}}{{{\text{Total number of cases in sample space}}}} \\\ = \dfrac{7}{{128}} \;$
So, the correct answer is “Option A”.

Note : It is interesting to note that we have excluded TTT from the favorable number of conditions as at the moment three simultaneous tails come then, the sequence of tossing will stop. Students must observe the given information in question carefully as exactly three tails have been asked in the question.