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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.

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Presentation on theme: "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."— Presentation transcript:

1 Practice ITCS 2175

2 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

3 How many ways can we get exactly 2 heads and 2 tails? 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 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 = Probability of k successes in n independent Bernoulli trials! C(n,k)p k q n-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 1 success: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: 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 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(E 1  E 2 ) = p(E 1 ) + p(E 2 ) - p(E 1  E 2 ) = 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.


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