30 questions · timed · auto-graded
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
$\text{None of these}$
The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is:
$\frac{7}{64}$
$\frac{7}{128}$
$\frac{45}{1024}$
$\frac{7}{41}$
Which one is not a requirement of a binomial dstribution?
$\big(\frac{9}{10}\big)^5$
$\frac{9}{10}$
$10^{-5}$
$\big(\frac{1}{2}\big)^2$
A coin is tossed 4 times. The probability that at least one head turns up is:
$\frac{1}{16}$
$\frac{2}{16}$
$\frac{14}{16}$
$\frac{15}{16}$
A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:
$\frac{1}{5}$
$\frac{1}{5}\big(\frac{9}{10}\big)^3$
$\big(\frac{3}{5}\big)^4$
$\text{None of these}$
In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is 3. Then, its mean is:
For a binomial variate X, if $\text{n}=3$ and $\text{P(X}=1)=8\text{ P(X = 3}),$ then p =
$\frac{4}{5}$
$\frac{1}{5}$
$\frac{1}{3}$
$\frac{2}{3}$
A coin is tossed n times. The probability of geting at least once is greater than 0.8. Then, the least value of n, is:
A coin is tossed 10 times. The probability of getting exactly six heads is:
$\frac{512}{513}$
$\frac{105}{512}$
$\frac{100}{153}$
$\text{ }^{10}\text{C}_6$
Solution:
$\text{n}=10,\text{x}=6,\text{p = q}=\frac{1}{2}$
$\text{P(X}=6)=\text{ }^{10}\text{C}_6\big(\frac{1}{2}\big)^{10}=\frac{105}{512}$
If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is:
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{10}\big(\frac{3}{4}\big)^6$
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)\big(\frac{3}{4}\big)^6$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)^6\big(\frac{3}{4}\big)^6$
Solution:
A fair die is thrown then probebility of getting 6 isp $=\frac{1}{6}.$
$\Rightarrow\text{q}=\frac{5}{6}$
To find probability that on tenth throw 4th six appears, in the first nine throw 3 six should appear.
Required probability = P(3 six in first 9 throw) × P(a six in tenth throw)
Required probability $=\text{ }^9\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^6\times\frac{1}{6}$
Required probability $=\frac{84\times5^6}{6^{10}}$
Solution:
X represents number of males.
$\text{p = q}=\frac{1}{2}$
$\text{p(n}-1)=\frac{3}{2^{10}}$
$\text{ }^{\text{n}}\text{C}_{\text{n}-1}\text{p}^{\text{n}-1}\text{q}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}-1}\big(\frac{1}{2}\big)^{\text{n}}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=\frac{1}{4}\times\frac{3\times4}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=12\big(\frac{1}{2}\big)^{12}$
$\Rightarrow\text{n}=12$
Solution:
$\text{n = 8,}\text{p}=\frac{1}{2}=\text{q}$
$\text{P(|X}-4|)\leq2$
$\Rightarrow-2\leq\text{x}-4\leq2$
$\Rightarrow4-2\leq\text{x}\leq2+4$
$\Rightarrow2\leq\text{x}\leq6$
$\text{P}(2\leq\text{x}\leq6)=\text{P(2)+P(3)+P(4)+P(5)+P(6)}$
$\text{P(2}\leq\text{x}\leq6)=\text{ }^8\text{C}_2\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_3\Big(\frac{1}{2^8}\Big)\\+\text{ }^8\text{C}_4\Big(\frac{1}{2^8}\Big)\text{ }^8\text{C}_5\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_6\Big(\frac{1}{2^8}\Big)$
$=\frac{119}{128}$