M.C.Q (1 Marks)

Question 11 Mark

If X is a binomial variate with parameters n and p, where 0 < p < 1 such that $\frac{\text{P(X = r)}}{\text{P(X = n - r})}$ is independent of n and r, then p equals:

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
$\text{None of these}$

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Question 21 Mark

Mark the correct alternative in the following question:The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is:

$\frac{7}{64}$
$\frac{7}{128}$
$\frac{45}{1024}$
$\frac{7}{41}$

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Question 31 Mark

Mark the correct alternative in the following question:Which one is not a requirement of a binomial dstribution?

There are 2 outcomes for each trial.
There is a fixed number of trials.
The outcomes must be dependent on each other.
The probability of success must be the same for all the trials.

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Question 41 Mark

In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective?

$\big(\frac{9}{10}\big)^5$
$\frac{9}{10}$
$10^{-5}$
$\big(\frac{1}{2}\big)^2$

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Question 51 Mark

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:

$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{10}\big(\frac{3}{4}\big)^6$
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)\big(\frac{3}{4}\big)^6$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)^6\big(\frac{3}{4}\big)^6$

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Question 61 Mark

A rifleman is firing at a distant target and has only 10% chance of hiting it. the least number of round he must fire in order to have more than 50% chance of hitting it at least once is:

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Question 71 Mark

A coin is tossed 4 times. The probability that at least one head turns up is:

$\frac{1}{16}$
$\frac{2}{16}$
$\frac{14}{16}$
$\frac{15}{16}$

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Question 81 Mark

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:

$\frac{1}{5}$
$\frac{1}{5}\big(\frac{9}{10}\big)^3$
$\big(\frac{3}{5}\big)^4$
$\text{None of these}$

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Question 91 Mark

In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is 3. Then, its mean is:

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Question 101 Mark

For a binomial variate X, if $\text{n}=3$ and $\text{P(X}=1)=8\text{ P(X = 3}),$ then p =

$\frac{4}{5}$
$\frac{1}{5}$
$\frac{1}{3}$
$\frac{2}{3}$

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Question 111 Mark

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:

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Question 121 Mark

A coin is tossed 10 times. The probability of getting exactly six heads is:

$\frac{512}{513}$
$\frac{105}{512}$
$\frac{100}{153}$
$\text{ }^{10}\text{C}_6$

Answer

$\frac{105}{512}$

Solution:
$\text{n}=10,\text{x}=6,\text{p = q}=\frac{1}{2}$
$\text{P(X}=6)=\text{ }^{10}\text{C}_6\big(\frac{1}{2}\big)^{10}=\frac{105}{512}$

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Question 131 Mark

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is:

$\frac{2}{3}$
$\frac{4}{5}$
$\frac{7}{8}$
$\frac{15}{16}$

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Question 141 Mark

The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8, is:

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Question 151 Mark

A fair coin is tossed 100 times. The probability of getting tails an odd nimber of times is:

$\frac{1}{2}$
$\frac{1}{8}$
$\frac{3}{8}$
$\text{None of these}$

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Question 161 Mark

Mark the correct alternative in the following question:
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P(X = r})}{\text{P(X = n} -\text{r})}$ is independent of n and r, then p equals:

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{5}$
$\frac{1}{7}$

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Question 171 Mark

If X follows a binomial distribution with parameter $\text{n}=100$ and $\text{p}=\frac{1}{3},$ then P(X = r) is maximum when r =

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Question 181 Mark

One hundred idential coins, each with probability p of showing heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, the value of p is:

$\frac{1}{2}$
$\frac{51}{101}$
$\frac{49}{101}$
$\text{None of these}$

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Question 191 Mark

A box contain 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens draws one by one with replacement at most one is defective?

$\big(\frac{9}{10}\big)^5$
$\frac{1}{2}\big(\frac{9}{10}\big)^4$
$\frac{1}{2}\big(\frac{9}{10}\big)^5$
$\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$

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Question 201 Mark

A biased coin with probabilty p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5},$ then p equals:

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{2}{5}$
$\frac{3}{5}$

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Question 211 Mark

Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9 is:

$\big(\frac{3}{7}\big)^7$
$\big(\frac{1}{15}\big)^7$
$\big(\frac{8}{15}\big)^7$
$\text{None of these}$

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Question 221 Mark

A fair die is tossed eight times. The probability that a third six is observed in the eight throw is:

$\frac{\text{ }^7\text{C}_2\times5^5}{6^7}$
$\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$
$\frac{\text{ }^7\text{C}_2\times5^5}{6^6}$
$\text{None of these}$

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Question 231 Mark

A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is:

$\frac{15}{2^8}$
$\frac{2}{15}$
$\frac{15}{2^{13}}$
$\text{None of these}$

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Question 241 Mark

A fair coin is tossed 99 times. If X is the number of times head appears, then P(X = r) is maximum when r is:

49, 50
50, 51
51,,52
None of these

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MCQ 251 Mark

A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is:

A
$\frac{\text{ }^{20}\text{C}_{10}\times5^6}{6^{20}}$
B
$\frac{120\times5^7}{6^{10}}$
✓
$\frac{84\times5^6}{6^{10}}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{84\times5^6}{6^{10}}$

A fair die is thrown then probebility of getting $6$ is $p =\frac{1}{6}.$$\Rightarrow\text{q}=\frac{5}{6}$
To find probability that on tenth throw $4^{th}$ six appears, in the first nine throw $3$ six should appear.
Required probability $= P(3$ six in first $9$ throw$) \times P($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 $=\frac{84\times5^6}{6^{10}}$

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Question 261 Mark

If in a binomial distribution $\text{n}=4,\text{P(X}=0)=\frac{16}{81},$ then $\text{P(X}=4)$ equals:

$\frac{1}{16}$
$\frac{1}{81}$
$\frac{1}{27}$
$\frac{1}{8}$

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Question 271 Mark

The probablity of selecting a male or a female is same. If the probability that in an office of n persons (n - 1) males being selected is $\frac{3}{2^{10}},$ the value of n is:

Answer

Solution:
X represents number of males.
$\text{p = q}=\frac{1}{2}$
$\text{p(n}-1)=\frac{3}{2^{10}}$
$\text{ }^{\text{n}}\text{C}_{\text{n}-1}\text{p}^{\text{n}-1}\text{q}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}-1}\big(\frac{1}{2}\big)^{\text{n}}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=\frac{1}{4}\times\frac{3\times4}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=12\big(\frac{1}{2}\big)^{12}$
$\Rightarrow\text{n}=12$

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Question 281 Mark

If X follows a binomial distribution with parameter $\text{n}=8$ and $\text{p}=\frac{1}{2},$ then $\text{P(|X}-4|\leq2)$ equals:

$\frac{118}{128}$
$\frac{119}{128}$
$\frac{117}{128}$
$\text{None of these}$

Answer

$\frac{119}{128}$

Solution:
$\text{n = 8,}\text{p}=\frac{1}{2}=\text{q}$
$\text{P(|X}-4|)\leq2$
$\Rightarrow-2\leq\text{x}-4\leq2$
$\Rightarrow4-2\leq\text{x}\leq2+4$
$\Rightarrow2\leq\text{x}\leq6$
$\text{P}(2\leq\text{x}\leq6)=\text{P(2)+P(3)+P(4)+P(5)+P(6)}$
$\text{P(2}\leq\text{x}\leq6)=\text{ }^8\text{C}_2\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_3\Big(\frac{1}{2^8}\Big)\\+\text{ }^8\text{C}_4\Big(\frac{1}{2^8}\Big)\text{ }^8\text{C}_5\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_6\Big(\frac{1}{2^8}\Big)$
$=\frac{119}{128}$

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Question 291 Mark

Mark the correct alternative in the following question:
The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is:

$\text{ }^5\text{C}_4(0.7)^4(0.3)$
$\text{ }^5\text{C}_1(0.7)(0.3)^4$
$\text{ }^5\text{C}_4(0.7)(0.3)^4$
$(0.7)^4(0.3)$

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Question 301 Mark

Let X denote the number of times heads occur in n tosses of a fair coin. If P(X = 4), P(X = 5) and P(X = 6) are in AP, the value of n is:

7, 14
10, 14
12, 7
14, 12

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MATHS M.C.Q (1 Marks)

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