Toss a fair coin 43 times what is the number of cases where number of heads number of tails

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A coin is tossed 10 times. The...

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Answer : B

Solution : <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NDA_14Y_MATH_C21_E01_043_S01.png" width="80%"> <br> Hence, total number of points in the sample space is 11.

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Hint: Suppose there are n number of tosses of fair coins. First of all, we will find the probability of getting 0 heads. That is the probability of getting the results of all the fair coin tosses as tails. Then we find the probability of getting only 1 head. The probability of not getting at least two heads will be equal to the sum of probability of getting no heads and probability of getting only 1 head out of n tosses. Thus, we can find the probability of not getting at least two heads. The probability of getting at least two heads will be one minus probability of not getting at least two heads. To find the probabilities, we will be making use of concepts of combinations.

Complete step-by-step answer:
The probability of getting a head in fair toss is $\dfrac{1}{2}$ and probability of getting a tail is $\dfrac{1}{2}$.
Let the number of tosses of a fair coin be n.
Let A be the event of getting only tails and no heads.
So, all the tosses will have tails simultaneously. Thus, the probability of n number of tails will be multiplied.
Thus, P(A) = $\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}...\dfrac{1}{2}$
$\Rightarrow $ P(A) = $\dfrac{1}{{{2}^{n}}}$
Let B be the event of getting only 1 head out of n tosses.
To get the probability, we need to choose one toss out of n tosses. Thus, number of ways of choosing 1 toss out of n tosses is $^{n}{{C}_{1}}$.
Thus, probability of getting one of the tosses as head and remaining tails will be given as:
$\Rightarrow $ P(B) = $^{n}{{C}_{1}}\times \dfrac{1}{2}\times \dfrac{1}{{{2}^{n-1}}}$
$\Rightarrow $ P(B) = $\dfrac{^{n}{{C}_{1}}}{{{2}^{n}}}$.
Let S be the event of not getting at least two heads. Thus, the probability of S will be equal to sum of probability A happening and probability of B happening. This is so because if either of the event happens event S will be true.
$\Rightarrow $ P(S) = $\dfrac{1}{{{2}^{n}}}+\dfrac{^{n}{{C}_{1}}}{{{2}^{n}}}$
$\Rightarrow $ P(S) = $\dfrac{1{{+}^{n}}{{C}_{1}}}{{{2}^{n}}}$
Therefore, S’ will be the event of getting at least two heads from n tosses.
$\Rightarrow $ P(S’) = $1-\dfrac{1{{+}^{n}}{{C}_{1}}}{{{2}^{n}}}$
$\Rightarrow $ P(S’) = $1-\dfrac{1+n}{{{2}^{n}}}$
It is given that the probability of S’ must be at least 0.96
$\Rightarrow $ P(S’) $\ge $ 0.96
$\begin{align}
  & \Rightarrow 1-\dfrac{1+n}{{{2}^{n}}}\ge 0.96 \\
 & \Rightarrow \dfrac{1+n}{{{2}^{n}}}<1-0.96 \\
 & \Rightarrow \dfrac{1+n}{{{2}^{n}}}<0.04 \\
\end{align}$
Now, we will check for various values of n, such that the inequality $\dfrac{1+n}{{{2}^{n}}}<0.04$.
If n = 5, $\dfrac{1+5}{{{2}^{5}}}=0.1875$
If n = 7, $\dfrac{1+7}{{{2}^{7}}}=0.0625$
If n = 8, $\dfrac{1+8}{{{2}^{8}}}=0.0351$
Therefore, the minimum number of tosses will be 8.

Note: If two events two or more events are happening simultaneously, the probability of all the event happening is given by the product of individual probability, whereas if of them happening or another is happening, then the probability of first or second event happening is given by the sum of probability of first event and probability of second event. To summarise, probability of A and B is P(A).P(B) and probability of A or B is P(A) + P(B).

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