4. Solve below: A. A fair coin is tossed 13 times. What is the probability that

Question

4. Solve below:

A.

A fair coin is tossed 13 times. What is the probability that the coin lands head at least 11 times?

a) 0.0111

b) 0.0225

c) 0.0112

d) 0.0095

e) 0.0110

f) None of the above.

B.

A fair coin is tossed 25 times. What is the probability that at most 24 heads occur?

a) 0.00000003

b) 0.99999997

c) 0.00000077

d) 0.00000075

e) 0.99999923

f) None of the above.

C.

A fair coin is tossed 29 times. What is the probability that at least 2 tails occur?

a) 0.99999995

b) 0.00000076

c) 0.99999919

d) 0.00000000

e) 0.99999994

f) None of the above.

Explanation / Answer

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 13 * 0.5
= 6.5
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 13 * 0.5 * 0.5
= 3.25
III.
standard deviation = sqrt( variance ) = sqrt(3.25)
=1.8028
A
P( X < 11) = P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(x=0)
= ( 13 10 ) * 0.5^10 * ( 1- 0.5 ) ^3 + ( 13 9 ) * 0.5^9 * ( 1- 0.5 ) ^4 + ( 13 8 ) * 0.5^8 * ( 1- 0.5 ) ^5 + ( 13 7 ) * 0.5^7 * ( 1- 0.5 ) ^6 + ( 13 6 ) * 0.5^6 * ( 1- 0.5 ) ^7 + ( 13 5 ) * 0.5^5 * ( 1- 0.5 ) ^8 + ( 13 4 ) * 0.5^4 * ( 1- 0.5 ) ^9 + ( 13 3 ) * 0.5^3 * ( 1- 0.5 ) ^10 + ( 13 2 ) * 0.5^2 * ( 1- 0.5 ) ^11 + ( 13 1 ) * 0.5^1 * ( 1- 0.5 ) ^12 + ( 29 0 ) * 0.5^0 * ( 1- 0.5 ) ^29
= 0.9888
P( X > = 11 ) = 1 - P( X < 11) = 0.0112
When P=0.5, the probability that the coin lands head at least 11 times = 0.0112
ANSWER:OPTION C
B.
P( X = 25 ) = ( 25 25 ) * ( 0.5^25) * ( 1 - 0.5 )^0
= 0.00000003
P( X < = 24) = 1 - P( X = 25 ) = 1 - 0.00000003
= 0.99999997
ANSWER: b) 0.99999997
C.
P( X < 2) = P(X=1) + P(X=0) +   
= ( 29 1 ) * 0.5^1 * ( 1- 0.5 ) ^28 + ( 29 0 ) * 0.5^0 * ( 1- 0.5 ) ^29 +
= 0.000000056
P( X > = 2 ) = 1 - P( X < 2) = 0.999999944
When fair coin is tossed 29 times. the probability that at least 2 tails occur is 0.999999944
ANSWER: OPTION e) 0.99999994

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