Question

Each letter of the word "supercalifragilisticexpialidocious" is placed into a bag and drawn at 3 times, replacing the letter after each draw. Find the probability that the letter "i" is drawn at least once.

Answer 1

Answer:

P(X ≥ 1) = 0.50

Step-by-step explanation:

Given that:

The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.

Then; the probability of success = 7/34 = 0.20588

Using Binomial distribution to determine the probability; we have:

$P(X = x) = ^nC_x \ \beta^x \ (1 - \beta)^{n-x}$

where;

x = 0,1,2,...n    and    0  <  β   <   1

and x represents the  number of successes.

However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:

P(X ≥ 1) = 1 - P(X< 1)

P(X ≥ 1) = 1 - P(X =0)

$P(X \ge 1) = 1 - \bigg [ {^3C__0} (0.21)^0 (1-0.21)^{3-0} \bigg]$

$P(X \ge 1) = 1 - \bigg [ 1 \times 1 (0.79)^{3} \bigg]$

P(X ≥ 1) = 1 - 0.50

P(X ≥ 1) = 0.50