Question

In a box there are 4 black beads and 3 red beads. In another box there
are 3 black beads and 5 red beads. If one bead each is taken from the
boxes, at random, what is the probability of getting,
a. Beads of the same colours
b. Beads of different colours
c. At least one black bead.

Answer 1

Step-by-step explanation:

box 1: 4 b, 3 r total beads=7

box 2: 3 b, 5 r total beads = 8

a)same beads

no. of ways of picking same color from

box 1= 4c2+3c2=9

box 2= 3c2+5c2=3+10=13

total no. of ways = 7c2+8c2 = 21+28=49

probability of selecting box 1=1/2, box 2=1/2

p(same beads)= 1/2×9/49 + 1/2×13/49 = 22/2×49=11/49

b)different beads

ways to pick different colors from

box 1 = 4c1×3c1=12

box 2 = 3c1×5c1=15

total no of ways = 49

p(different beads) = 1/2×12/49+1/2×15/49=27/2×49 =27/98

c)atleast one black

ways to pick non black from

box 1 = 3c2=3

box 2= 5c2 =10

at least one black = 1- non black

p(atleast one black) = 1 -( 1/2×3/49 +1/2×10/49) = 1-(13/2×49)= 98-13/98 = 85/98

i tried to solve this with my knowledge in probability. hope this helps you