Question

Round all answers to 4 decimal places. a. A bag contains 6 green marbles, 10 red marbles, and 3 blue marbles. If a marble is drawn from the bag. replaced and another marble is drawn, what is the probability of drawing first a green marble and then a blue marble? 0.8421 * b. A bag contains 10 white marbles, 3 blue marbles, and 6 red marbles. If two different marbles are drawn from the bag. What is the probability of drawing first a white marble and then a red marble?

Answer 1

Groeen Marloles Real = l0 t1 Total madle= 6 +10+3 The bebabilit ta dra fba Green maole aa thin elue mas ble 3 z O-0498 X witt plece) o.o499 hite = 10+3+ 6 Blus To tn 19 Probabil i osing fivat a white end thin LO X O166 2

Phebabiliy eingfit a uhta and red th n maxbles witautapl 8 O-1754 17 19x18 의드
my dear student the condition for (b) is not given so i have solved two ways.with replacement .and without replacement.Thank you

Answer 2

SOLUTION :

a.

Total marbles in the bag = 6 + 10 + 3 = 19

P(first green)

= Number of green marbles in the bag / Total marbles in there bag

= 6/19 

It is replaced in the bag, so total marbles in the bag is 19 as earlier and having same composition.

So,

P(2nd blue)

= Number of blue marbles / Total marbles

= 3/19

So,

P(1st green, 2nd blue with replacement)

= 6/19 * 3/19

= 0.0500 (ANSWER).

b.

Total balls = 10 + 3 + 6 = 19

P(1st white marble)

= Number of white marbles / Total marbles

= 10/19

@nd ball is drawn (without replacement). 

So before 2nd draw :

Number of white marbles = 10 - 1 = 9 ; 3 blue marbles ; 6 red marbles.

Total marbles = 9 + 3 + 6 = 18

So,

P(2nd red | 1st white which is not replaced)

= Number of red marbles just before 2nd draw / Total marbles at that time

= 6/18

= 1/3 

Hence,

P(1st white, 2nd red without replacement)

= P(1st white) * P(2nd red | 1st white which is not replaced)

= 10/19 * 1/3

= 10/57   (ANSWER)