If an unbiased coin is tossed n times. Find the probability that head appears an odd number of times.

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 succeses is

• 28/256

• $\frac{219}{256}$

• 128/256

• 37/256

An unbiased coin is tossed n times. Let X denotes the number of times head occurs. If P(X = 4 ), P(X = 5) and PX = 6) are in AP, then the value of n can be

• 7, 14

• 10, 14

• 12, 7

• None of these

Let X and Y be two events such that P(X/Y) = 1/2, P(Y/X) = 1/3 and P(X ∩ Y) = 1/6. Which of the following is incorrect?

• P(X $\cup$ Y) = 2/3

• X and Y are independent

• P(X^C ∩ Y) = 2/3

• X and Y are not independent

If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is

• $\frac{4^{\mathrm{n}}}{5^{\mathrm{n}}}$

• $\frac{2^{\mathrm{n}}}{5^{\mathrm{n}}}$

• 4 - $\frac{2^{\mathrm{n}}}{5^{\mathrm{n}}}$

• None

A bag contains (n + 1) coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is $\frac{7}{12}$, then the value of n is

• 5

• 4

• 3

• None of these

If A and B are two given events, then P(A ∩ B) is

• equal to P(A) + P(B)

• equal to P(A) + P(B) + P(A $\cup$ B)

• not less than P(A) + P(B) - 1

• not greater than P(A) + P(B) - P(A $\cup$ B)

From a city population, the probability of selecting a male or smoker $\frac{7}{10}$, a male smoker is $\frac{2}{5}$ and a male, if a smoker is already, selected, is $\frac{2}{3}$ Then, the probability of

• selecting a male is $\frac{3}{2}$

• selecting a smoker is $\frac{1}{5}$

• selecting a non-smoker is $\frac{2}{5}$

• selecting a smoker, if a male is first selected, is given by $\frac{8}{5}$

If A, B, C are three events associated with a random experiment, then $\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{A}} \cap \mathrm{B}\right)$

• $\mathrm{P}\left(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}\right)$

• $\mathrm{P}\left(\mathrm{A}\hspace{0.17em}\cap \mathrm{B}\hspace{0.17em}\cap \mathrm{C}\right)$

• $\mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{A}} \cap \mathrm{B}\right)$

• $\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)$

1150.

The probability of atleast one double six being thrown in n throws with two ordinary dice is greater than 99%.

Then, the least numerical value of n is

• 100

• 164

• 170

• 184