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Mean= E(X) =
= 0×0.3+1×0.7
=0.7
= 02×0.3+12×0.7 =0.7
Now, ∴ Var(X)= E(X2)(E(X))2
= 0.7−(0.7)2
= 0.7 − 0.49
= 0.21
A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of successes is:
In a single toss, P(success) = 2/6 = 1/3 and P(nonsuccess) = (1 − 1/3) = 2/3.
P(X=0) = P(nonsuccess in the 1st draw and nonsuccess in the second)
(2/3 × 2/3) = 4/9.
P(X=1) = P(success in the 1st toss and nonsuccess in the 2nd) or (nonsuccess in the 1st toss and success in the 2nd)]
(1/3 x 2/3) + (2/3 × 1/3) = 4/9.
A man make attempts to hit the target. The probability of hitting the target is 3/5 . Then the probability that A hit the target exactly 2 times in 5 attempts is:
Probability of success(p) = ⅗
q = 1  p
= 1  (⅗) ⇒ ⅖
Probability of 2 hit in 5 attempts = 5C2 (⅗)^{2} (⅖)^{3}
= 144/625
Trials of a random experiment are called Bernoulli trials, if they satisfy the condition/s:
In binomial probability distribution, mean is 3 and standard deviation is 3/2 . Then the probability distribution is:
Mean = np = 3, S.D. (npq)^{½} = 3/2
p = ¼ n = 12
Hence, binomial distribution is (p+q)^{n} = (¾ + ¼)^{12}
A random variable is a real valued function whose domain is the.
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
Probability of getting first nondefective bulb =90/100
Probablity of getting second nondefective bulb =89/99
Probablity of getting third nondefective bulb =88/98
Probablity of getting fourth nondefective bulb =87/97
Probablity of getting fifth nondefective bulb =86/96
So, probablity of getting all nondefective bulbs in a sample of 5 bulbs =
(90/100)∗(89/99)∗(88/98)∗(87/97)∗(86/96)
= closest option is b
The mean and variance of a binomial distribution are 4 and 3 respectively, then the probability of getting exactly six successes in this distribution is:
Let X˜B(n,p) be a binomial variate with mean 4 and variance 3. Then,
np=4,and npq=3
⇒q=3/4,p=1/4 and n=16
∴P(X=r)=.16Cr(1/4)^{r }(3/4)^{(16−r)} ,r=0,1,2,....,16
⇒P(X=6)= 16C6(1/4)^{6} ((3)/(4))^{10}
In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb is put back in the box before making the second selection. The probability that both the bulbs are not defective is:
Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. The probability that there is at least one defective egg is:
Probability of an egg being defective =10/100=110
So, probability of an egg being nondefective=1−0.1=0.9
10 eggs are drawn successively with replacement.
So, the probability of getting no defective egg =(0.9)^{10}
Hence, the probability that there is at least one defective egg = 1−(0.9)^{10}
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