Suppose an urn has six balls, three red, two blue, and one yellow. On any draw from the urn, all balls in the urn are equally likely to be drawn. Suppose that two balls are drawn, in sequence without replacement, that is, the second ball is drawn without replacing the first ball in the urn. What is the probability that the second ball drawn is blue

1 / 3

Step-by-step explanation:

Number of balls :

Red, R = 3 ; Blue, B = 2 ; Yellow, Y = 1

Total = 6

From two draws without replacement, probability that 2nd draw is blue :

Possibilities :

P(RB) or P(YB) or P(BB)

P(RB) = 3/6 * 2/5 = 6/30

P(YB) = 1/6 * 2/5 = 2 /30

P(BB) = 2/6 * 1/5 = 2 / 30

HENCE,

P(RB) + P(YB) + P(BB)

6/30 + 2/30 + 2/30 = 10/30 = 1/3

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )