# How do I solve a problem consisting of independent events?

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

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

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 $P(X\ge2)$.

• 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 Aug 19, 2016 at 12:58
• 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. Aug 19, 2016 at 13:01
• thank you for your time. i wont be gone Aug 19, 2016 at 13:02
• @Utsav I've edited my answer to make things clearer. Aug 19, 2016 at 13:53

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 $${3\choose2}\cdot \left({3\over10}\right)^2{7\over10}+{3\choose 3}\left({3\over10}\right)^2={27\over125}=0.216\ .$$

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}