Zozina

Newest questions and responses by Zozina
  1. Probability

    Question:A fair coin is flipped independently until the first Heads is observed. Let the random variable K be the number of tosses until the first Heads is observed plus 1. For example, if we see TTTHTH, then K=5. For K=1,2,3...K, let Xk be a continuous

    asked on October 27, 2018
  2. Probability: Counting

    Hi everyone. I am struggling a bit with this one example question I am doing. Problem: There is a group of 12 people, 6 men and 6 women. A committee is to be formed consisting of 5 members from this group. Find the probability that Anne (one of the

    asked on September 24, 2018
  1. math

    Yes. By definition, the mode is the number that occurs most frequently in the set. Therefore it has to be in the set.

    posted on November 19, 2018
  2. Math

    How many letters are there in the alphabet? 26 How many vowels are there in the alphabet? 5 Therefore, the probability of picking one vowel is 5/26. The probability of picking N vowels is (5/26)^N or 5/26 * 5/26 * 5/26.... (N times). So for your question,

    posted on November 19, 2018
  3. Math

    @Help Yes that is the right answer.

    posted on November 19, 2018
  4. Probability

    Sorry, typo it should be (R+1)^2 there, not (R+1)^1

    posted on November 18, 2018
  5. Probability

    @Anonymous, E[((R+1)^1 - (R-1)^2)/100] = E[4R/100] =1/25 * E[R] = 1/25 * 3.5 = 0.14.

    posted on November 18, 2018
  6. Probability

    1) The random circle can only intersect the circle of radius 5 if the centre of the random circle is at most 1 unit away from the circumference of the circle of radius 5. So basically we can draw two boundaries above and below the circle of radius 5. These

    posted on November 18, 2018
  7. Probability

    Ok, I know my mistake, I didn't read the question carefully, but was on the right track

    posted on October 27, 2018
  8. Probability: Counting

    Oh I simply divide those two results I mentioned, but I am not sure why that works. Because can't, Anne and Billy, be placed in the committee in different ways? 10C3 / 12C5 makes it seem like the ways in which Billy and Anne can be arranged in the

    posted on September 25, 2018
  9. Probability: Counting

    Hi Scott. Hmm, I am still unsure what you mean by that. I guess the probability of choosing Anne, in particular, is 1/12, and then the probability of choosing Billy after that becomes 1/11. So maybe 1/12 * 1/11 * (10C3)/(12C5) ? I am quite confused.

    posted on September 24, 2018