A bag contains 6 identical balls numbered 1,2,3,4,5 and 6.
Two balls are drawn at random, one after another, from the bag w/o
replacement.
(i ) Draw the possiblity diagram to show the outcome of the
draw.
(ii) Find, as a fraction in its simplest form, the probability
that
(a) both balls have an even no.
(b) the sum of the numbers drawn is 8
(c) the product of the numbers drawn is 7
(d) at least one of the numbers is a multiple of 3
(i) Draw probability tree
(ii)
(a) 3/6 * 2/5 = 1/5
(b) Sum is 8, you have (2,6), (3,5) (5,3) (6,2)
So, 1/6 * 1/5 * 4 choices = 2/15
(c) 7 is a prime number, so only factors of 1 and 7
Probability is zero
(d) Ans = 1 - P(no mulitple of 3) = 1 - 4/6 * 3/5 = 3/5
Do help confirm... I did it mental and posted it
