its an easy question

just use idiot, i mean 5!/2!

multiply eash fragment by 3,3,1
54/125 + 36/125 + 8/125 = .784

Sana, you just took the Patern as 1B2W without considering the different possible patterns formed by them.
For example:
BWW
WBW
WWB
So, you need to multiply Step 1 & 2 for Second Way by 3 to get the answer

My Steps:

3(4 x 6 x 6)/1000 + 3(4 x 4 x 6)/1000 + (4 x 4 x 4)/1000
= 784/1000 = .784

sana.noor wrote:i am trying to solve this question in two different ways, one way is easy and i get right answer but the other way is lengthy but i am not getting the right answer, am i doing it right? here is the question and explanation
there are 100 balls in a jar. 40 are black and 60 are white. what is the probability of getting at least one black in three consecutive picks with replace of each ball.
a) .216
b).316
c).5
d).684
e).784
OA is 0.784

one way: 1-(getting no black ball) = 1- (3/5)^3= 0.784 answer

2nd way: 1) getting one black and 2 white= 18/125
2)getting two black and one white = 12/125
3) getting 3 black = 8/125

adding all three cases = 38/125 =.304 and thats the wrong answer, what i am doing wrong here

Probability of getting at least 1 black ball = 1 - probability of getting no black balls
= 1 - (6/10)(6/10)(6/10)
= 1 - (0.6 * 0.6 * 0.6)
= 1 - 0.216
= 0.784

Choose e

The other way:
p(one black ball) = (4/10)(6/10)(6/10) * 3C1____________________3C1 because the order matters
= 0.144*3
= 0.432
p(2 black balls) = (4/10)(4/10)(6/10) * 3C2
= 0.096*3
= 0.288
p(3 black balls) = (4/10)(4/10)(4/10)
= 0.064

So p(at least one black ball) = 0.432 + 0.288 + 0.064 = 0.784

Choose e

Cheers

sana.noor wrote:i am trying to solve this question in two different ways, one way is easy and i get right answer but the other way is lengthy but i am not getting the right answer, am i doing it right? here is the question and explanation
there are 100 balls in a jar. 40 are black and 60 are white. what is the probability of getting at least one black in three consecutive picks with replace of each ball.
a) .216
b).316
c).5
d).684
e).784
OA is 0.784

one way: 1-(getting no black ball) = 1- (3/5)^3= 0.784 answer

1) getting one black and 2 white= 18/125
2) getting two black and one white = 12/125
3) getting 3 black = 8/125

adding all three cases = 38/125 =.304 and thats the wrong answer, what i am doing wrong here

Case 1: exactly 1 black marble
You have correctly determined P(BWW):
The probability that the FIRST marble is black, while the second and third marbles are white.
But this is not the ONLY way to get exactly 1 black marble.
We must account for ALL of the ways to get exactly 1 black marble.
Since the 1 black marble could be selected on any of the 3 picks -- BWW, WBW, WWB -- we multiply by 3:
3 * 18/125 = 54/125.

Case 2: exactly 2 black marbles, implying exactly 1 white marble
You have correctly determined P(BBW):
The probability that the FIRST AND SECOND MARBLES are black, while the third marble is white.
But this is not the ONLY way to get exactly 2 black marbles.
We must account for ALL of the ways to get exactly 2 black marbles.
Since the 1 white marble could be selected on any of the 3 picks -- WBB, BWB, BBW -- we multiply by 3:
3 * 12/125 = 36/125.

Case 3: all 3 marbles are black
You have correctly determined P(BBB) = 8/125.
Since there is only one way to get exactly 3 black marbles -- BBB -- we don't need to multiply here.

Adding the fractions, we get:

54/125 + 36/125 + 8/125 = 98/125 = .784.

thanks everyone!