1 / 8

Practice

Practice. ITCS 2175. Flipping a coin 4 times. How many different combinations of heads and tails are there? C1: HHHHHHHH T T T T T T T T C2: HHHHT T T THHHH T T T T C3: HH T T H H T THH T THHT T C4: HT H TH T H TH T HT H THT 16: How do we compute without counting? Product rule: 2 4.

lilly
Download Presentation

Practice

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Practice ITCS 2175

  2. Flipping a coin 4 times. • How many different combinations of heads and tails are there? C1: HHHHHHHHTTTTTTTT C2: HHHHTTTTHHHHTTTT C3: HHTTHHTTHHTTHHTT C4: HTHTHTHTHTHTHTHT 16: How do we compute without counting? Product rule: 24

  3. How many ways can we get exactly 2 heads and 2 tails? C1: HHHHHHHHTTTTTTTT C2: HHHHTTTTHHHHTTTT C3: HHTTHHTTHHTTHHTT C4: HTHTHTHTHTHTHTHT 6: How do we compute? C(4,2) = 4!/2!(4-2)! = 24/4 = 6 Why is this a combination and not a permutation?

  4. What is the probability of getting exactly 2 heads? 6/16 = 0.375 Probability of k successes in n independent Bernoulli trials! C(n,k)pkqn-k = C(4,2)(0.5)2(0.5)2 = 6*(0.5)4 =0.375

  5. What is the probability of getting at least one tails in 4 flips? C1: HHHHHHHHTTTTTTTT C2: HHHHTTTTHHHHTTTT C3: HHTTHHTTHHTTHHTT C4: HTHTHTHTHTHTHTHT 15/16 1success:C(4,1)(1/2)3(1/2)1 = (4!/3!1!)/16 = 4/16 2 successes: C(4,2 )(1/2)2(1/2)2 = (4!/2!2!)/16 = 6/16 3 successes: C(4,3 )(1/2)2(1/2)2 = (4!/1!3!/16 = 4/16 4 successes: C(4,4)(1/2)2(1/2)2 = (4!/4!0!)/16 = 1/16 0 successes: C(4,0)(1/2)4(1/2)0 = (4!/0!4!)/16 = 1/16

  6. What is the probability of getting a tails on the 2nd flip? C1: HHHHHHHHTTTTTTTT C2: HHHHTTTTHHHHTTTT C3: HHTTHHTTHHTTHHTT C4: HTHTHTHTHTHTHTHT 8/16 = 1/2 There are two possibilities on the 2nd flip, heads or tails. Therefore, the probability of tails is 1/2. Each flip is an independent event!

  7. What is the probability of getting a tails on either the 2nd flip or the 4th flip? C1: HHHHHHHHTTTTTTTT C2: HHHHTTTTHHHHTTTT C3: HHTTHHTTHHTTHHTT C4: HTHTHTHTHTHTHTHT 12/16 = 3/4 p(tails on 2nd flip) + p(tails on 4th flip) - p(tails on both flips) = 8/16 + 8/16 - 4/16 = 3/4 p(E1E2) = p(E1) + p(E2) - p(E1E2) = 1/2 + 1/2 - 1/4 = 3/4

  8. What is the probability that we have a heads immediately followed by at least one tails with 4 coin flips? C1: HHHHHHHHTTTTTTTT C2: HHHHTTTTHHHHTTTT C3: HHTTHHTTHHTTHHTT C4: HTHTHTHTHTHTHTHT 12/16 : How to compute? Three places that can be a heads 1/2 is probability that a heads is there, 1/2 is probability that tails follows: 3*(1/2)*(1/2) = 3/4.

More Related