NIMCET 2024 — Mathematics PYQ

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Question

NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

A coin is thrown 8 number of times. What is the probability of getting a head in an odd number of throw?

Choose the correct answer:

Explanation

When a fair coin is tossed, the probability of getting a head ( $p$ ) and the probability of getting a tail ( $q$ ) in any single throw are:

$p = \frac{1}{2}$

$q = \frac{1}{2}$

The coin is thrown $n = 8$ times. We need to find the probability of getting a head an odd number of times. This means we want to find the probability of getting exactly 1, 3, 5, or 7 heads.

Using the Binomial Distribution Formula, the probability of getting exactly $r$ successes in $n$ trials is given by:

$P (X = r) = (r n) \cdot p^{r} \cdot q^{n - r}$

Substituting $n = 8$ , $p = \frac{1}{2}$ , and $q = \frac{1}{2}$ :

$P (X = r) = (r 8) \cdot (\frac{1}{2})^{r} \cdot (\frac{1}{2})^{8 - r} = (r 8) \cdot (\frac{1}{2})^{8}$

Step 1: Write the expression for the total probability of odd outcomes

$P (odd number of heads) = P (X = 1) + P (X = 3) + P (X = 5) + P (X = 7)$

$P (odd number of heads) = [(1 8) + (3 8) + (5 8) + (7 8)] \cdot (\frac{1}{2})^{8}$

Step 2: Use Binomial Identity properties

According to the properties of binomial coefficients, the sum of odd binomial coefficients is equal to half of the total sum of all binomial coefficients:

$(1 n) + (3 n) + (5 n) + \dots = 2^{n - 1}$

For $n = 8$ :

$(1 8) + (3 8) + (5 8) + (7 8) = 2^{8 - 1} = 2^{7}$

Step 3: Calculate the final probability value

Substitute this value back into our probability equation:

$P (odd number of heads) = 2^{7} \cdot (\frac{1}{2})^{8}$

$P (odd number of heads) = \frac{2 ^{7}}{2 ^{8}}$

$P (odd number of heads) = \frac{1}{2}$