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(a). \[\dfrac{{63}}{{64}}\]

(b). \[\dfrac{1}{2}\]

(c). \[\dfrac{{127}}{{128}}\]

(d). \[\dfrac{{255}}{{256}}\]

Answer

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Hint: Find the number of outcomes where we don’t obtain at least one head or at least one tail, that is, when the outcomes are only tail or only head respectively. Then, subtract from total outcomes and find the probability.

Complete step-by-step answer:

Let E be the event of obtaining at least one head and at least one tail.

Since, the coin is unbiased, the head and the tail appear with equal probability each time it is tossed and hence, this is an equally likely event.

Let the total number of outcomes be N(S) and the number of outcomes favourable to event E be N(E), then the probability P(E) of obtaining at least one head and at least one tail is given as follows:

\[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{N\left( E \right)}}{{N\left( S \right)}}{\text{ }}.........{\text{(1)}}\]

The total number of outcomes by tossing a coin eight times is given as follows:

\[N(S) = 2.2.2.2.2.2.2.2{\text{ (eight times)}}\]

\[N(S) = {2^8}\]

\[N(S) = 256{\text{ }}.........{\text{(2)}}\]

Number of outcomes of at least one head and at least one tail can also be found by subtracting the number of outcomes, where at least one head and at least one tail do not occur, from the total number of outcomes, that is:

\[N(E) = N(S) - N(\bar E){\text{ }}............{\text{(3)}}\]

Here, \[\bar E\] is the event where at least one head and at least one tail don’t occur. In other words, it is the event where only the tail or only head occurs.

The number of outcomes where only the head (HHHHHHHH) appears is one and the number of outcomes where only the tail (TTTTTTTT) appears is one. Then the number of outcomes where only head or only tail occurs is the sum of these two.

\[N(\bar E) = 1 + 1\]

\[N(\bar E) = 2{\text{ }}...........{\text{(4)}}\]

Substituting equation (2) and equation (4) in equation (3), we get:

\[N(E) = 256 - 2{\text{ }}\]

\[N(E) = 254{\text{ }}...........{\text{(5)}}\]

Now, substituting equation (2) and equation (5) in equation (1), we get the probability as follows:

\[P(E) = \dfrac{{254}}{{256}}\]

Simplifying further, we obtain:

\[P(E) = \dfrac{{127}}{{128}}\]

Hence, the correct answer is option (c).

Note: You may proceed to find the number of outcomes of at least one head and at least one tail directly case by case but that is not the correct method since it is time consuming and complicated. Instead, find the cases where only head or only tail occurs and subtract from the total number of cases, which is the right method to do.

Complete step-by-step answer:

Let E be the event of obtaining at least one head and at least one tail.

Since, the coin is unbiased, the head and the tail appear with equal probability each time it is tossed and hence, this is an equally likely event.

Let the total number of outcomes be N(S) and the number of outcomes favourable to event E be N(E), then the probability P(E) of obtaining at least one head and at least one tail is given as follows:

\[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{N\left( E \right)}}{{N\left( S \right)}}{\text{ }}.........{\text{(1)}}\]

The total number of outcomes by tossing a coin eight times is given as follows:

\[N(S) = 2.2.2.2.2.2.2.2{\text{ (eight times)}}\]

\[N(S) = {2^8}\]

\[N(S) = 256{\text{ }}.........{\text{(2)}}\]

Number of outcomes of at least one head and at least one tail can also be found by subtracting the number of outcomes, where at least one head and at least one tail do not occur, from the total number of outcomes, that is:

\[N(E) = N(S) - N(\bar E){\text{ }}............{\text{(3)}}\]

Here, \[\bar E\] is the event where at least one head and at least one tail don’t occur. In other words, it is the event where only the tail or only head occurs.

The number of outcomes where only the head (HHHHHHHH) appears is one and the number of outcomes where only the tail (TTTTTTTT) appears is one. Then the number of outcomes where only head or only tail occurs is the sum of these two.

\[N(\bar E) = 1 + 1\]

\[N(\bar E) = 2{\text{ }}...........{\text{(4)}}\]

Substituting equation (2) and equation (4) in equation (3), we get:

\[N(E) = 256 - 2{\text{ }}\]

\[N(E) = 254{\text{ }}...........{\text{(5)}}\]

Now, substituting equation (2) and equation (5) in equation (1), we get the probability as follows:

\[P(E) = \dfrac{{254}}{{256}}\]

Simplifying further, we obtain:

\[P(E) = \dfrac{{127}}{{128}}\]

Hence, the correct answer is option (c).

Note: You may proceed to find the number of outcomes of at least one head and at least one tail directly case by case but that is not the correct method since it is time consuming and complicated. Instead, find the cases where only head or only tail occurs and subtract from the total number of cases, which is the right method to do.

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