Question

3) A box contains 10 tickets bearing only one number from 1 to 10 on
each. If one ticket is drawn at random, find the probability of an
event that the ticket drawn is an odd number​

Answer 1

Answer:

Step-by-step explanation:

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}  ∴ n(S) = 30  i. Let A be the event that the ticket drawn bears an odd number.  ∴ A = {1,3,5,7,9,11,13,15,17,19,21, 23,25,27,29} ∴ n(A) = 15 ∴ P(A) = ()() n ( A ) n ( S )  = 15/30 ∴  P(A) = 1/2 ii. Let B be the event that the ticket drawn bears a complete square number.  ∴ B = {1,4,9,16,25}  ∴ n(B) = 5 ∴ P(B) = ()() n ( B ) n ( S ) = 5/30 ∴ P(B) = 1/6 ∴ P(A) = 1/2; P(B) = 1/6Read more on Sarthaks.com - https://www.sarthaks.com/847594/contains-tickets-bearing-only-one-number-from-each-ticket-drawn-random-find-probability

Answer 2

Answer:

the answer is 5/10=1/2 ....