Question

A bag contains 4 white,5 blacks and 2 blue balls. 3 balls are drawn one after the other without replacement from the bag .what is the probability that they are of different color​

Answer 1

Answer:

24/99

Step-by-step explanation:

From the question given above, the following data were obtained:

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

Probability that they are of different color =?

Next, we shall determine the total number of balls in the bag. This can be obtained as follow:

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

TOTAL = 4 + 5 + 2 = 11 balls

Next, we shall determine the possible outcome of draw. This can be obtained as follow.

The possible outcome could be:

WBLB or WBBL or BBLW or BWBL or BLBW or BLWB

Next we shall determine the probability of each of outcomes.

Since the ball is drawn without replacement, it means the total number of ball will reduce after each draw.

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

TOTAL = 11

P(WBLB) = 4/11 × 2/10 × 5/9 = 40/990

P(WBLB) = 4/99

P(WBBL) = 4/11 × 5/10 × 2/9 = 40/990

P(WBBL) = 4/99

P(BBLW) = 5/11 × 2/10 × 4/9 = 40/990

P(BBLW) = 4/99

P(BWBL) = 5/11 × 4/10 × 2/9 = 40/990

P(BWBL) = 4/99

P(BLBW) = 2/11 × 5/10 × 4/9 = 40/990

P(BLBW) = 4/99

P(BLWB) = 2/11 × 4/10 × 5/9 = 40/990

P(BLWB) = 4/99

Finally, we shall determine the probability that they are of different color. This can be obtained as follow:

P(WBLB) = 4/99

P(WBBL) = 4/99

P(BBLW) = 4/99

P(BWBL) = 4/99

P(BLBW) = 4/99

P(BLWB) = 4/99

Probability that they are of different color =?

Probability that they are of different color = P(WBLB) + P(WBBL) + P(BBLW) + P(BWBL) + P(BLBW) + P(BLWB)

= 4/99 + 4/99 + 4/99 + 4/99 + 4/99 + 4/99

= (4 + 4 + 4 + 4 + 4 + 4)/99

= 24/99

Thus, the probability that they are of different color is 24/99