Three tickets are selected at random from a box of tickets bearing numbers from 1 to 30. the table and the relative-frequency histogram show the distribution of the number of even-numbered tickets chosen. find the given probability. p(2 odd-numbered tickets) a. 13/116 b. 15/116 c. 45/116 d. 71/116

Sample space = { 15 odd + 15 even} =30 possible outcomes
In how many ways I can choose 3 tickets: ³⁰C₃
In how many ways I can choose 2 tickets with ODD number : ¹⁵C₂
In how many ways I can choose 1 ticket with EVEN number : ¹⁵C₁

The probability of getting 2 odds & 1 even (3 tickets are drawn) is

P( probability choosing 2 ODDS & 1 EVEN) = (¹⁵C₂ x ¹⁵C₁) / ³⁰C₃ = 45/116

29933333.33

step-by-step explanation:

6×10=604000000÷60=66666.6666766666.6667×449=29933333.33