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Question: A leather bag contains 4 black beads, 3 red beads and three white beads. Inside a plastic bag are 5 black beads, 2 red beads and 3 white beads. Another nylon bag contains 6 black beads, 1 red bead and 3 white beads. One bead is randomly withdrawn from each bag.

What is the probability of getting at least two white beads?

My workout:

1- (4/5 * 4/5 * 4/5) # the probability of not getting 2 white beads and the only way for that to happen is if only 1 white bead is picked therefore leaving 8 to pick from. And since each bag has 10 beads including 3 white beads the chance for this is 8/10= 4/5 This is my logic behind this

My final answer:

1- (4/5 * 4/5 * 4/5) = 1 - 64/125 = 61/125

My problem with this now is, I don't know if this is the answer or I have missed a step. Any help will be appreciated

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3 Answers 3

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We are drawing one bead each from the three bags, and we are drawing them independently. There is a $\frac3{10}$ chance of drawing a white bead from each bag. If we get at least two white beads, we could have got them from

  • the nylon and plastic bags
  • the leather and nylon bags
  • the leather and plastic bags
  • all three bags

For each of the first three cases it is implied that we draw a non-white bead from the third bag, which has a $\frac7{10}$ chance of happening. The probability of each of these cases happening is therefore $\frac3{10}\times\frac3{10}\times\frac7{10}=\frac{63}{1000}$; we multiply this by three for the probability of getting exactly two white beads, which works out to be $\frac{189}{1000}$.

Similarly, the last case (drawing exactly three white beads) has probability $\frac3{10}\times\frac3{10}\times\frac3{10}=\frac{27}{1000}$ of occurring. Since drawing two white beads and drawing three white beads are mutually exclusive events, add them together to get your answer: $\frac{189}{1000}+\frac{27}{1000}=\frac{216}{1000}=\frac{27}{125}$.

If you look a little deeper this is really a binomial distribution $X$ with $n=3$ and $p=\frac3{10}$; we have just calculated $P(X\ge2)$.

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  • $\begingroup$ I have just started on this topic and your way of understanding is very complex for me to take in. Isn't there another way of understanding this better? thanks $\endgroup$
    – Utsav
    Aug 19, 2016 at 12:58
  • $\begingroup$ Basically, your reasoning is wrong. There are three independent events of drawing a bead, each of which has a 3/10 chance of returning a white bead. So if at least two whites are drawn, I could draw them from the nylon/plastic bags, the leather/nylon bags, the leather/plastic bags or all three - and I sum them together. I'll add a clarification quite soon - don't go away yet. $\endgroup$ Aug 19, 2016 at 13:01
  • $\begingroup$ thank you for your time. i wont be gone $\endgroup$
    – Utsav
    Aug 19, 2016 at 13:02
  • $\begingroup$ @Utsav I've edited my answer to make things clearer. $\endgroup$ Aug 19, 2016 at 13:53
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There are three independent trials with success probability ${3\over10}$ for each. In such a case the number of successes is binomially distributed. The probability of at least two successes then comes to $${3\choose2}\cdot \left({3\over10}\right)^2{7\over10}+{3\choose 3}\left({3\over10}\right)^2={27\over125}=0.216\ .$$

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My workout:

1- (4/5 * 4/5 * 4/5) # the probability of not getting 2 white beads and the only way for that to happen is if only 1 white bead is picked therefore leaving 8 to pick from. And since each bag has 10 beads including 3 white beads the chance for this is 8/10= 4/5 This is my logic behind this

No that the probability of not drawing a white bead with a black mark given that you so marked only two of the three white beads in each bag.

"At least two" is "not one nor none".   That is, if we let $W$ be the count of white marbles drawn: $$\begin{align}\mathsf P(W\geq 2)~=~&1-\mathsf P(W=0)-\mathsf P(W=1)\\ =~& 1 - (\tfrac 7{10})^3 - 3(\tfrac 7{10})^2\tfrac 3{10} \\ =~& \tfrac{1000-343-441}{1000} \\ =~& \tfrac{216}{1000} \\ =~& \dfrac {27}{125}\end{align}$$

$\mathsf P(W=0)=\tfrac 7{10}\tfrac 7{10}\tfrac 7{10}$ as it is the probability of obtaining not-white from each bag.

$\mathsf P(W=1)= \tfrac 3{10}\tfrac 7{10}\tfrac 7{10}+\tfrac 7{10}\tfrac 3{10}\tfrac 7{10}+\tfrac 7{10}\tfrac 7{10}\tfrac 3{10} =3(\tfrac 7{10})^2\tfrac 3{10}$ as it is the probability of obtaining not-white from two bags and white from the other in any of the $3$ orders the white may emerge.

Alternatively

$$\begin{align}\mathsf P(W\geq 2)~=~&\mathsf P(W=2)+\mathsf P(W=3)\\ ~=~& 3(\tfrac 3{10})^2\tfrac 7{10} + (\tfrac 3{10})^3 \\ ~=~& \dfrac {27}{125}\end{align}$$

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