1
$\begingroup$

Could I get help on this "sampling without replacement" question please?

In a drawer, there are 10 socks, six RED and four BLUE.

Q1: If you were to get four socks in a row, what is the probability that the first two are RED and the last two are BLUE?

Q2: On the same drawer (with the 10 socks in it), you get four socks in a row, what is the probability that exactly two are red?

I though that Q1 would be: 6/10 * 5/9 30/90 = 0.33

(Six out of 10, then five out of 9 - since we already got one of the six reds). And the probability of blue would be: 4/8 * 3/7 = 0.5 * 0.428 = 0.21 Then the total probability is the sum of both = 54%

But I'm wrong. And I have no idea how to solve for the second.

Can I have some guidance please? (I know that this can be solved with the combination formula, but I'd like to understand the probabilities, if possible)

$\endgroup$
  • $\begingroup$ please add the self-study tag, and read its tag wiki, modifying your question as necessary (in particular, you'll need to be more specific about what help you need -- a good deal more specific than "can I have some guidance" and "I have no idea"). $\endgroup$ – Glen_b Jan 18 '16 at 22:53
4
$\begingroup$

You almost got that right!

What you need to remember, is that when you want the probability of two events happenning, you need to multiply the probability of both event to happen. If A is the first event and B the second, the probability that both event happen, $p(A \text{ and } B) = p(A)p(B)$ (if they are independent).

When you want the probability that either one or the other to happen, you need to add their probability; $p(A \text{ or } B) = p(A) + p(B)$.

One way to remember it is as follow: Probabilites are between 0 and 1. If you multiply two probabilities, the result is smaller than each of the initial probabilities. This is what happens when you require two events to happen; it is less likely that both of them happen than each of them individually. If you add them, the result is bigger. This is what happens when you want either one or the other, the probability that one or the other happens is bigger than each of them.

For Q1, it is an and problem. The chain of events that needs to happen RRBB (for two red (R), two blue (B) socks).

  • The probability that the first is R is $6/10$.
  • The probability that the second is R, knowing that the first one was R, is $5/9$.
  • To get the probability that both events happenend (R and R), you need to multiply them; The probability that you pick RR is $6/10 \times 5/9$.
  • Then it continues; the probability that the third one is B, knowing that the first two are R, is $4/8$.
  • To get the probability that RRB happened, all three events need to happen, so it is $5/10 \times 4/9 \times 4/8$. I think you can manage the last sock from there.

For Q2, it is an or problem; You take 4 socks and want to know the probability that exactly two of them are red. This could happen from the following scenarios;

  • Scenario S1: RRBB
  • Scenario S2: RBRB
  • Scenario S3: RBBR
  • Scenario S4: BRRB
  • Scenario S5: BRBR
  • Scenario S6: BBRR

Since you want the probability that either S1 or S2 or S3 ... or S6, you need to compute $p(S1) + p(S2) + p(S3) + ... + p(S6)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.