In a large school district, it is known that 25% of all students entering kindergarten are already reading. A simple random sample of 10 new kindergarteners is drawn. What is the probability that fewer than three of them are able to read?

A. 2503

B. 2816

C. 5000

D. 5256

E. 7759

Any formulas or ways to remember when to use a certain formula would be great, seeing as how I forgot how to do this stuff!

Given Information:

Probability of success = p = 25% = 0.25

Number of trials = n = 10

Required Information:

P (x < 3) = ?

Answer:

P (x < 3) = 0.5256 (option D)

Step-by-step explanation:

The given problem can be solved using binomial distribution since:

There are n repeated trials and are independent of each other. There are only two possibilities: student is able to read or student is not able to read. The probability of success does not change with trial to trial.

The binomial distribution is given by

P (x) = ⁿCₓ pˣ (1 - p) ⁿ⁻ˣ

Where n is the number of trials, x is the variable of interest and p is the probability of success.

The variable of interest in this case is fewer than 3 students and the probability that fewer than three students are able to read is

P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)

For P (x = 0):

Here we have x = 0, n = 10 and p = 0.25

P (x = 0) = ¹⁰C₀ (0.25⁰) (1 - 0.25) ¹⁰⁻⁰

P (x = 0) = (1) (0.25⁰) (0.75) ¹⁰

P (x = 0) = 0.0563

For P (x = 1):

Here we have x = 1, n = 10 and p = 0.25

P (x = 1) = ¹⁰C₁ (0.25¹) (1 - 0.25) ¹⁰⁻¹

P (x = 1) = (10) (0.25¹) (0.75) ⁹

P (x = 1) = 0.1877

For P (x = 2):

Here we have x = 2, n = 10 and p = 0.25

P (x = 2) = ¹⁰C₂ (0.25²) (1 - 0.25) ¹⁰⁻²

P (x = 2) = (45) (0.25²) (0.75) ⁸

P (x = 2) = 0.28157

Finally,

P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)

P (x < 3) = 0.0563 + 0.1877 + 0.28157

P (x < 3) = 0.52557

Rounding off yields

P (x < 3) = 0.5256

Therefore, the probability that fewer than three of the students are able to read is 0.5256