Question Details

A fair coin is tossed 20 times. The probability that 'head' will appear exactly 4 times in the first ten tosses, and ‘tail’ will appear exactly 4 times in the next ten tosses is ______ (round off to 3 decimal places)

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Correct Answer :

Correct answer is : 0.042

Probability of 4 heads in first 10 tosses and 4 tails in next ten tosses will be given by

P (4 head in 10 tosses) × P (4 tail in 10 tosses)

Probability of 4 heads in first 10 tosses

n = 10, , r = 4

P (4 head in 10 tosses) = ¹⁰C₄ × 0.5⁴ × 0.5⁶ = 0.205

Probability of 4 tails in last 10 tosses

n = 10, , r = 4

P (4 tail in 10 tosses) = ¹⁰C₄ × 0.5⁴ × 0.5⁶ = 0.205

∴ Probability of 4 heads in first 10 tosses and 4 tails in next ten tosses = 0.205 × 0.205 = 0.0420

Solution :

The correct answer is 0.042.

To find the probability of getting exactly 4 heads in the first 10 tosses and exactly 4 tails in the next 10 tosses, we can analyze the two sets of 10 tosses independently. Since each coin toss is an independent event, the outcome of the first ten tosses does not affect the outcome of the next ten tosses.

Therefore, the joint probability is the product of the individual probabilities:
P(exactly 4 heads in first 10 tosses AND exactly 4 tails in next 10 tosses) = P(4 heads in first 10 tosses) × P(4 tails in next 10 tosses)

Let us calculate each probability using the binomial probability formula:
$P (X = k) = (\binom{n}{k} p^{k} q^{n - k}$
where:
n = 10 (number of trials/tosses in each set)
k = 4 (number of successful outcomes)
p = 0.5 (probability of success on a single toss for a fair coin)
q = 1 - p = 0.5 (probability of failure on a single toss)

Step 1: Probability of getting exactly 4 heads in the first 10 tosses
Here, success is getting a head (p = 0.5) and failure is getting a tail (q = 0.5).
$P (4 heads in 10 tosses) = (\binom{10}{4}) \times {0.5}^{4} \times {0.5}^{6}$
First, compute the binomial coefficient:
$(\binom{10}{4}) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$
Now substitute this back into the formula:
$P (4 heads in 10 tosses) = 210 \times {0.5}^{10} = 210 \times 0.0009765625 \approx 0.2051$

Step 2: Probability of getting exactly 4 tails in the next 10 tosses
Using the same logic, where success is getting a tail (p = 0.5), we get:
$P (4 tails in 10 tosses) = (\binom{10}{4}) \times {0.5}^{4} \times {0.5}^{6} \approx 0.2051$

Step 3: Calculating the combined probability
Multiply the two independent probabilities together:
$Total Probability = 0.2051 \times 0.2051 \approx 0.04206$
Rounding off to 3 decimal places, we get:
Total Probability = 0.042