probability

Question

5. A fair coin is tossed four times.
a. List the entire sample space for this experiment.
Use your sample space from part (a) to answer the following questions:
b. What is the probability that it turns up heads every time?
c. What is the probability of at least 2 heads?
d. What is the probability of two tails in a row?

Answer 1

a) Listing all 16 possible sequences:

HHHH  HHHT  HHTH  HHTT

HTHH  HTHT  HTTH   HTTT

THHH  THHT  THTH   THTT

TTHH  TTHT  TTTH    TTTT

 

b) probability of 4 heads =(1/2)^4 or 1/16 or 6.25%   (probability of a single event raised to the number of trials).

      as can be seen above only one out of 16 sequences has 4 heads

 

c) probability of getting at least 2 heads, this is a cumulative probability, it equals the probability of getting 2 heads, 3 heads or 4 heads so three binomial equations must be solved

b(x;n,p) = nCx * p^x * (1-p)^(n-x) where n is the number of trials, x is the number of successes and p is the                                                                  probability of success for an individual trial,

 

nCx is equal to the combinations and is equal to:n!/[x!(n-x)!] 

so for two heads exactly: in this case n=4, x=2 and p = 0.5

b(x;n,p) = nCx * p^x * (1-p)^(n-x) =  4!/[2!(4-2)!] * (0.5)^2 * (1-0.5)^(4-2) = 0.375 

 

so for three heads exactly: in this case n=4, x=3 and p = 0.5

b(x;n,p) = nCx * p^x * (1-p)^(n-x) = 4!/[3!(4-3)!] * (0.5)^3 * (1-0.5)^(4-3) = 0.25

 

so for four heads exactly: in this case n=4, x=4 and p = 0.5

b(x;n,p) = nCx * p^x * (1-p)^(n-x) = 4!/[4!(4-4)!] * (0.5)^4 * (1-0.5)^(4-4) = 0.0625

so the total probability is 0.375 + 0.25 + 0.0625 = .6875 or 68.75%

 

d)  probability of getting 2 tails in a row , this again is a cumulative probability, it equals the probability of getting 2 tails, or 3 tails in a row out of the sequence of 4 rolls taken 2 at a time so two binomial equations must be solved. 

 

b(x;n,p) = nCx * p^x * (1-p)^(n-x) where n is the number of trials, x is the number of successes and p is the probability of success for an individual trial, nCx is equal to the combinations and is equal to:n!/[x!(n-x)!] 

 

so for two tails in a row: in this case n=3, x=2 and p = 0.5
b(x;n,p) = nCx * p^x * (1-p)^(n-x) = 3!/[2!(3-2)!] * (0.5)^2 * (1-0.5)^(3-2) = 0.375

so for three tails in a row: in this case n=3, x=3 and p = 0.5
b(x;n,p) = nCx * p^x * (1-p)^(n-x) = 3!/[3!(3-3)!] * (0.5)^3 * (1-0.5)^(3-3) = 0.125

so the total probability = 0.375 + 0.125 or 50%

 

 

 

 

 

 

 

 

Answer 2

Sample space equals: 

 

HHHH

HHHT

HHTH

HTHH

THHH

TTTT

TTTH

TTHT

THTT

HTTT

THHT

HTTH

HHTT

TTHH

THTH

HTHT

 

b.) 1/16

c.) 11/16

d.) 5/16