The probability that a page of book contains a typo is 0.2 and only one typo may occur in a page. If 13 pages of that book are selected, ?

P(a page contains a typo) = 0.2
P(a page does not contain typo) = 1 - 0.2 = 0.8
x = number of typos in 13 pages

P(2 ≤ x ≤ 5)
= P(x =2) + P(x = 3) + P(x = 4) + P(x = 5)
= ₁₃C₂(0.2)²(0.8)¹¹ + ₁₃C₃(0.2)³(0.8)¹⁰ + ₁₃C₄(0.2)⁴(0.8)⁹ + ₁₃C₅(0.2)⁵(0.8)⁸
= 0.2680 + 0.2457 + 0.1535 + 0.0691
= 0.7363

The answer: A. 0.7363

Binomial Distribution Probability:

P(x) = n! / [(n - x)! x!] p^x (1 - p)^(n - x)

Where:

x = number of times you want a positive result (2, 3, 4, and 5)
n = number of events run total (13)
p = probability of a positive result per test (0.2)

Substitute what we know except for x to get our function, then solve for:

P(2) + P(3) + P(4) + P(5)

P(x) = 13! / [(13 - x)! x!] 0.2^x * 0.8^(13 - x)

P(2) = 13! / [(13 - 2)! 2!] 0.2^2 * 0.8^(13 - 2)
P(2) = 13! / (11! * 2) 0.2^2 * 0.8^11
P(2) = 13 * 12 / 2 * 0.04 * 0.08589934592
P(2) = 0.26801 (rounded to 5DP)

P(3) = 13! / [(13 - 3)! 3!] 0.2^3 * 0.8^(13 - 3)P(3) = 13! / (10! * 6) 0.2^2 * 0.8^10P(3) = 13 * 12 * 11 / 6 * 0.008 * 0.1073741824P(3) = 0.24567 (rounded to 5DP)P(4) = 13! / [(13 - 4)! 4!] 0.2^4 * 0.8^(13 - 4)P(4) = 13! / (9! * 24) 0.2^4 * 0.8^9P(4) = 13 * 12 * 11 * 10 / 24 * 0.0016 * 0.134217728P(4) = 0.15355 (rounded to 5DP)P(5) = 13! / [(13 - 5)! 5!] 0.2^5 * 0.8^(13 - 5)
P(5) = 13! / (8! * 120) 0.2^5 * 0.8^8
P(5) = 13 * 12 * 11 * 10 * 9 / 120 * 0.00032 * 0.16777216
P(5) = 0.06910

Now add the results together and round to 4 DP:

0.26801 + 0.24567 + 0.15355 + 0.06910
0.7363 (rounded to 4DP)

Answer A.