Fischer and Spassky play a chess match in which the first player to win a game wins the match. After 10 successive draws, the ma
1 answer:
Answer:
a)
the probability that Fischer wins the match is 0.5714
b)
p( D = d ) = {
, d = 1, 2, 3, ........
otherwise
Step-by-step explanation:
Given that;
probability that Fischer wins the match, p = 4
probability that Spassky wins the match, q = 0.3
Match drawn, 1 - p - q = 1 - 0.4 - 0.3 = 0.7
(a) What is the probability that Fischer wins the match?
P( Fischer wins) = p/( p+q)
we substitute
P( Fischer wins) = 0.4 / ( 0.4 + 0.3)
P( Fischer wins) = 0.4 / 0.7
P( Fischer wins) = 0.5714
Therefore, the probability that Fischer wins the match is 0.5714
b) What is the PMF of the duration of the match?
let D represent the duration of the match
since the duration D of the match is a geometric random variable with parameter p + q,
the PMF will be;
p( D = d ) = ( p + q )
= ( 0.4 + 0.3 )
p( D = d ) = ( 0.7)
that is, p( D = d ) = {
, d = 1, 2, 3, ........
otherwise
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Answer:
the probability of no defects in 10 feet of steel = 0.1353
Step-by-step explanation:
GIven that:
A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two.
Let consider β to be the average value for defecting
So;
β = 2
Assuming Y to be the random variable which signifies the anticipated number of defects in a 10-foot segment of this roll.
Thus, y follows a poisson distribution as number of defect is infinite with the average value of β = 2
i.e
the probability mass function can be represented as follows:
where;
y = 0,1,2,3 ...
Hence, the probability of no defects in 10 feet of steel
y = 0
P(y =0) = 0.1353