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Get Full Access to Probability And Statistical Inference - 9 Edition - Chapter 1.1 - Problem 4e
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# A fair coin is tossed four times, and the sequence of

ISBN: 9780321923271 41

## Solution for problem 4E Chapter 1.1

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.

= 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.

= {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.

= {HTTT, TTTH, TTHT, THTT}

=

=

Therefore, P(BD) = 0.25.

Step 2 of 3

Step 3 of 3