Question

the digits 2, 5, 6, 7, and 8 are randomly arranged to form a five - digit number. complete parts (a) and (b) below.
(a) find the probability that the number is odd.
the probability that the number is odd is
(type an integer or a simplified fraction.)
(b) find the probability that the first and last digits of the number both are odd.
the probability that the first and last digits of the number both are odd is
(type an integer or a simplified fraction.)

Explanation:

Step1: Determine total permutations

The digits are 2, 5, 6, 7, 8. Total number of five - digit numbers (permutations of 5 digits) is \(n = 5! = 5\times4\times3\times2\times1=120\).

Step2: Solve part (a) - Probability the number is odd

A number is odd if its last digit is odd. The odd digits among 2, 5, 6, 7, 8 are 5 and 7 (2 odd digits).

• If the last digit is fixed as an odd digit (2 choices), the remaining 4 digits can be permuted in \(4!\) ways. So the number of odd five - digit numbers \(m_a=2\times4! = 2\times24 = 48\).
• Probability \(P_a=\frac{m_a}{n}=\frac{48}{120}=\frac{2}{5}\).

Step3: Solve part (b) - Probability first and last digits are odd

• First, choose the first and last digits (both odd). The odd digits are 5 and 7. The number of ways to choose and arrange the first and last digits (permutations of 2 odd digits for first and last) is \(P(2,2)=\frac{2!}{(2 - 2)!}=2! = 2\) (using permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), here \(n = 2\), \(r=2\)).
• The middle 3 digits (from the remaining 3 digits) can be permuted in \(3!\) ways. So the number of five - digit numbers with first and last digits odd \(m_b=2!\times3! = 2\times6 = 12\).
• Probability \(P_b=\frac{m_b}{n}=\frac{12}{120}=\frac{1}{10}\).

Answer:

(a) \(\frac{2}{5}\)
(b) \(\frac{1}{10}\)