Each of urns I and II has 5 red balls, 3 white balls, and 2 green balls. Urn III has 1 red ball, 1 white ball, and 8 green balls
1 answer:
Answer:
The probability of selecting the green ball from urn III is ²/₃
Step-by-step explanation:
Given;
total number of balls in urn I = 10
number of green balls in urn I , = 2
total number of balls in urn II = 10
number of green balls in urn II , = 2
total number of balls in urn III = 10
number of green balls in urn II , = 8
Therefore, the probability of selecting the green ball from urn III is ²/₃
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If p(x) and q(x) are arbitrary polynomials of degree at most 2 then
||p||||q|| = 26() and angle between p(x) and q(x) is 0.233.
Given that
<p,q> = p(-1)q(-1) + p(0) q(0) + p(3)q(3)
and p(x) = 2x²+ 6 , q(x)= 4x²-4x
then the values of p and q at x = -1,0,3 are given as;
x = -1,
p(-1) = 2(-1)² + 6 = 8 , q(-1) = 4(-1)² - 4(-1) = 8
x = 0,
p(0) = 2(0)² + 6 = 6 , q(0) = 4(0)² - 4(0) = 0
x = 3,
p(3) = 2(3)² + 6 = 24 , q(3) = 4(3)²- 4(3) = 24.
<p,q> = p(-1)q(-1) + p(0)q(0) + p(3)q(3)
= (8)(8) + (6)(0) + 24(24)
= 64 + 0 + 576
<p,q> = 640
Now we have to find ||p|| ||q||, for this we'll find ||p|| and ||q||
||p|| =
=
=
||p|| = 26
and
||q|| =
=
||q|| =
∴||p||||q|| = 26()
Now we have to find angle between p(x) and q(x),
∴ α = cos⁻¹
= cos ⁻¹
= cos ⁻¹
α = 13.34°
In radian
α = 0.233.
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