A bag contains different colored beads. The probability of drawing two black beads from the bag without replacement is 3/35 , and the probability of drawing one black bead is 3/10 .

What is the probability of drawing a second black bead, given that the first bead is black?

RedBlue
Dec 21, 2017

#1**+2 **

P(Two Black Beads) = P(Black Bead on First Draw) * P(Black Bead on Second Draw)

So we have

(3/35) = (3/10) * P(Black Bead on Second Draw)

Multiply both sides by (10/3)

(10/3)(3/35) = P(Black Bead on Second Draw) =

10/35

We can confirm this

Let A = P(Black Bead on first Draw) = 3/10

Let B = P(Black Bead on the Second Draw)

Let P(A and B) = 3/35

So

P(B / A) = P(A and B) / P(A) = (3/35) / (3/10) = (3/35)(10/3) = 10/35

CPhill
Dec 21, 2017

#1**+2 **

Best Answer

P(Two Black Beads) = P(Black Bead on First Draw) * P(Black Bead on Second Draw)

So we have

(3/35) = (3/10) * P(Black Bead on Second Draw)

Multiply both sides by (10/3)

(10/3)(3/35) = P(Black Bead on Second Draw) =

10/35

We can confirm this

Let A = P(Black Bead on first Draw) = 3/10

Let B = P(Black Bead on the Second Draw)

Let P(A and B) = 3/35

So

P(B / A) = P(A and B) / P(A) = (3/35) / (3/10) = (3/35)(10/3) = 10/35

CPhill
Dec 21, 2017

#4**+3 **

Sir CPhill's method is correct.

\(P(B|A)={\dfrac {P(A\cap B)}{P(A)}} \\\\ \text { Solve for } P(A\cap B) =\dfrac{3}{35} \text{ and } P(A) =\dfrac{3}{10}\\ P(B|A)= \Large( \small \frac{3/10}{3/35} \Large) = \small \dfrac{2}{7}\\\)

His final answer has an arithmetic error.

Just an itsy-bitsy, teeny-weeny, microscopic, nearly infinitesimally small, but calculable ^{fúckup}!

I did that once, when I was a toddler chimp, because someone substituted plantains for some of the bananas I was counting.

GingerAle
Dec 21, 2017

#5**+3 **

Duh!!....10/35 = 2/7

[ I always tell Sisyphus to reduce ...unfortunately.....dieting isn't part of his regimen ]

Thanks, GA!!!!

CPhill
Dec 21, 2017

#6**0 **

GA: It looks like you effed up again! Maybe you should eat boiled plantains instead of bananas! Or maybe crow! Happy chowing!.

Guest Dec 21, 2017

#10**+2 **

Well, thank you for the catered meal, Mr. BB! Crow is a delicacy for us chimps, and plantains are a fine complement for such a meal, especially when boiled and reduced in blarney sauce. I usually dress for fine dining, so I will don my gay apparel and Troll the ancient Mr. BB’s carol: Troll-la-Troll-la-Troll, la-Troll-Troll, Troll!

Lancelot Link had Stu, and Nauseated had Sorasyn, but they left, taking their dumbness elsewhere. I have you, Mr. BB, the stubborn, relentless, intractable Blarney Banker of lore: a pseudo intellectual with a multiplicity of advanced dimwit degrees in arrogant stupìdity, a professor of misinformation, who teaches with authority and irritation.

One usually needs a scanning electron microscope to find an error by Sir CPhill, That’s why the errors are “itsy-bitsy, teeny-weeny, microscopic, nearly infinitesimally small,” but still calculable. For you, I need no such instrument. A casual stroll through the forum shows a path, clearly marked by stupìdity, mòronic analysis, and toxic misinformation, leading right to your hovel.

You may not believe this, but I’m glad you’re here. The most successful Trolls have a reliable parade of stubborn dùmb dùmbs, or a relentless, indefatigable dùmb dùmb who always gives the wrong answer to “What is the airspeed velocity of an unladen swallow?” No matter how many times my troll character tosses you over the bridge railing into the river, you always come back for more. With this, you have personified a modern attachment to the ancient legends of Camelot and its Trolls.

We all owe you a debt of pseudo gratitude for the unwitting humor you cause.

GA

GingerAle
Dec 22, 2017

#11**+4 **

A bag contains different colored beads. The probability of drawing two black beads from the bag without replacement is 3/35 , and the probability of drawing one black bead is 3/10 .

What is the probability of drawing a second black bead, given that the first bead is black?

The prob of drawing one black bead is 3/10 so

There are 3n black beads and 7n non-black beads where n is a positive integer making 10n altogether:

P(BB with no replacement) =

\(\frac{3}{10}\times \frac{3n-1}{10n-1}=\frac{3}{35}\\ \frac{3n-1}{10n-1}=\frac{3}{35}\times \frac{10}{3}\\ \frac{3n-1}{10n-1}=\frac{2}{7}\\ 21n-7=20n-2\\ n=5\)

So there is 15 black and 35 non-black and 50 altogether

If one black has already been selected then there will be 14 black and 49 altogether remaining.

So the probability of selecting a second black is \(\frac{14}{49}=\frac{2}{7}\)

I seem to have done it the long way but our answers are all the same. :/

Melody
Dec 23, 2017

#12

#13**+1 **

Thanks Chris,

I get confused with all 'Ginger's' set symbols. I should become more familiar with them I suppose.

My way is using basic algebra that most of us are more comfortable with.

Then again, your way is just using basic algebra too and it is much easier than what I did.

I like your approach much better than mine. :) I think Ginger's is the same as yours. :)

Melody
Dec 23, 2017