-5x^5+35x^4-30x^3 can someone help me pls
1 answer:
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Answer:
a) P (0 defective component) = 0.5
P( 1 defective component ) = 0.35
P( 2 defective component ) = 0.15
b) P( 0 ) = 0
p ( 1 ) = 1
p ( 2 ) = 0.33
Step-by-step explanation:
p( no defective ) = 0.5
p( 1 is defective ) = 0.35
p( 2 is defective ) = 0.15
Given that 2 components are selected at random
<u>a) Given that neither component is defective </u>
Probability of 0 defective component = 0.5
P( 1 defective component ) = 0.35
P( 2 defective component ) = 0.15
<u>b) Given that one of the two tested component is defective </u>
P( 0 defective ) = 0
P( 1 defective ) = P ( ) = p( x = 1 ) / 1 - P ( x = 0 )
= ( 0.5 )^1 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^1
= 0.5 * 1 / 1 - 0.5 = 0.5 / 0.5 = 1
p ( 2 defective ) = p( x = 3 ) / 1 - P ( x = 0 )
= ( 0.5 )^2 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^2
= 0.25 / 0.75 = 0.33
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Answer:
0.6804
Step-by-step explanation:
Given that Margie is practicing for an upcoming tennis tournament. Her first serve is good 20 out of 30 times on average.
Since each trial is independent and there are two outcomes, X no of good serves is binomial with n=6 and p =2/3
Required probability
= Prob that atleast four of 6 times good serve
=
=
Formula used:
P(X=r) =