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A fair coin is tossed four times, and the sequence of

Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman ISBN: 9780321923271 41

Solution for problem 4E Chapter 1.1

Probability and Statistical Inference | 9th Edition

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Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman

Probability and Statistical Inference | 9th Edition

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Problem 4E

PROBLEM 4E

A fair coin is tossed four times, and the sequence of heads and tails is observed.

(a) List each of the 16 sequences in the sample space S.

(b) Let events A, B, C, and D be given by A = {at least 3 heads}, B = {at most 2 heads}, C = {heads on the third toss}, and D = {1 head and 3 tails}. If the probability set function assigns 1/16 to each outcome in the sample space, find

Step-by-Step Solution:
Step 1 of 3

Solution 4E

Step1 of 3:

We have A fair coin is tossed four times, and the sequence of heads and tails is observed.

We need to find,

(a) List each of the 16 sequences in the sample space S.

(b) Let events A, B, C, and D be given by A = {at least 3 heads}, B = {at most 2 heads},

 C = {heads on the third toss}, and D = {1 head and 3 tails}. If the probability set function assigns 1/16 to each outcome in the sample space, find (i)P(A), (ii)P(AB),(iii)P(B),

(iv)P(AC), (v)P(D), (vi)P(AC), (vii)P(BD)

Step2 of 3:

a).

If A fair coins tosses four times, the following is the list of 16 sequences in the sample space S.

S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTTH, HTTT, HTHT, THHH, TTHH, THTH, THHT, TTTH, TTHT, THTT, TTTT}


b).

1).Consider,

    A = {at least 3 heads}

        = {HHHH, HHHT, HHTH, HTHH, THHH}

        = 5

    Now,

    P(A) =

             =

             = 0.3125

    Therefore, probability of at least 3 heads is 0.3125.

2).Consider,

    B = {at most 2 heads}

        = {HHHH, HHHT, HHTH, HHTT, HTHH, HTTH, HTHT, THHH, TTHH, THTH, THHT}

        = 11

    Now,

    P(B) =

             =

             = 0.6875

    Therefore, probability of at most 2 heads is 0.6875.

3).Consider,

    C = {heads on the third toss}

        = {HHHH, HHHT, HTHH, HTHT, THHH, TTHH, THHT, TTHT}

        = 8

    Now,

    P(C) =

             =

             = 0.5000

    Therefore, probability of heads on the third toss is 0.5000.

4).Consider,

    D = {1 head and 3 tails}

        = { HTTT, TTTH, TTHT, THTT}

        = 4

    Now,

    P(D) =

             =

             = 0.2500

    Therefore, probability of 1 head and 3 tails is 0.2500.

Step3 of 3:

(i).P(A) =

             =

             = 0.3125

     Therefore, probability of at least 3 heads is 0.3125.

(ii).P(AB) = P{(at least 3 heads)(at most 2 heads)}

                     = There is no possible cases that at least 3 heads and at most 2 heads

      Therefore, P(AB) = 0.

(iii).P(B) =

               =

               = 0.6875

    Therefore, probability of at most 2 heads is 0.6875.

(iv).P(AC) = P{(at least 3 heads)(heads on the third toss)}

                      = {HHHH, HHHT, HTHH, THHH}

                      =

                      =

       Therefore, P(AC) = 0.25.

(v).P(D) =

             =

             = 0.2500

      Therefore, probability of 1 head and 3 tails is 0.2500.

(vi).P(AC) = P(A) + P(C) - P(AC)

                      = 0.3125 + 0.50 - 0.25

                      = 0.5625

       Therefore, P(AC) = 0.5625.

(vii).P(BD) = P{(at most 2 heads)(1 head and 3 tails)}

                      = {HTTT, TTTH, TTHT, THTT}

                      =

                      =

       Therefore, P(BD) = 0.25.

 

 

 

Step 2 of 3

Chapter 1.1, Problem 4E is Solved
Step 3 of 3

Textbook: Probability and Statistical Inference
Edition: 9
Author: Robert V. Hogg, Elliot Tanis, Dale Zimmerman
ISBN: 9780321923271

The full step-by-step solution to problem: 4E from chapter: 1.1 was answered by , our top Statistics solution expert on 07/05/17, 04:50AM. This textbook survival guide was created for the textbook: Probability and Statistical Inference , edition: 9. The answer to “A fair coin is tossed four times, and the sequence of heads and tails is observed.(a) List each of the 16 sequences in the sample space S.(b) Let events A, B, C, and D be given by A = {at least 3 heads}, B = {at most 2 heads}, C = {heads on the third toss}, and D = {1 head and 3 tails}. If the probability set function assigns 1/16 to each outcome in the sample space, find” is broken down into a number of easy to follow steps, and 79 words. This full solution covers the following key subjects: heads, space, tails, sample, most. This expansive textbook survival guide covers 59 chapters, and 1476 solutions. Since the solution to 4E from 1.1 chapter was answered, more than 467 students have viewed the full step-by-step answer. Probability and Statistical Inference was written by and is associated to the ISBN: 9780321923271.

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