Probability
An unbiased coin is tossed $$5$$ times. Suppose that a variable $$X$$ is assigned the value $$k$$ when $$k$$ consecutive heads are obtained for $$k=3,4,5,$$ otherwise $$X$$ takes the value $$-1$$. The the expected value of $$X$$, is :
An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is.
$$p=P(H)=\dfrac {1}{2}$$, $$q=P(T)=1_{/2}$$
$$P$$ (at least one head & one tail)
$$P(1 < H < 7)=P(1 < H)$$
$$=1-[\displaystyle p(H)=0 $$ and $$ p(H)=8]$$
$$=1-\left[2\cdot (1{/2})^8\right]$$
$$=1-\left[2\cdot \displaystyle\frac{1}{16\times 16}\right]$$
$$=1-\displaystyle\frac{1}{128}$$
$$=\displaystyle\frac{127}{128}$$.
An unbiased coin is tossed $$5$$ times. Suppose that a variable $$X$$ is assigned the value $$k$$ when $$k$$ consecutive heads are obtained for $$k=3,4,5,$$ otherwise $$X$$ takes the value $$-1$$. The the expected value of $$X$$, is :
In a workshop, there are five machines and the probability of any one of them to be out of service on day is $$\dfrac{1}{4}$$. If the probability that at most two machines will be out of service on the same day is $$\left(\dfrac{3}{4} \right)^3 k$$, then k is equal to :
If the mean and the variance of a binomial variate X are 2 and 1 respectively, then probability that X takes a value greater than or equal to one is
In a binomial distribution $$\mathrm{B} ($$n,p $$=\displaystyle \dfrac{1}{4})$$, if the probability of at least one success is greater than or equal to $$\displaystyle \dfrac{9}{10}$$ , then $$\mathrm{n}$$ is greater than
The minimum number of times one has to toss fair coins so that probability of observing at least one head is at least $$90$$% is:
An experiment succeeds twice as often as it fails. The probability of at least $$1$$ successes in the six trials of this experiment is:
A multiple choice examination has $$5$$ questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get $$4$$ or more correct answers just by guessing is
Consider 5 independent Bernoulli trials each with probability of success $$\mathrm{p}$$. If the probability of at least one failure is greater than or equal to $$\displaystyle \frac{31}{32}$$, then $$\mathrm{p}$$ lies in the interval.
If in a binomial distribution $$n=4,P(X=0)=\cfrac { 16 }{ 81 } $$, then $$P(X=4)$$ equals
The odds in favour of India winning any cricket match is $$2 : 3$$. What is the probability that if India plays $$5$$ matches, it wins exactly $$3$$ of them?