Probability · Mathematics · KCET

A random experiment has five outcomes $\mathrm{w}_1, \mathrm{w}_2, \mathrm{w}_3, \mathrm{w}_4$ and $\mathrm{w}_5$. The probabilities of the occurrence of the outcomes $w_1, w_2, w_3, w_4$ and $w_5$ are respectively $\frac{1}{6}, a, b$ and $\frac{1}{12}$ such that $12 a+12 b-1=0$. Then the probabilities of occurrence of the outcome $w_3$ is

A die has two face each with number ' 1 ', three faces each with number ' 2 ' and one face with number ' 3 '. If the die is rolled once, then $\mathrm{P}(1$ or 3$)$ is

Consider the following statements.

Statement (I): If E and F are two independent events, then $E^{\prime}$ and $F^{\prime}$ are also independent.

Statement (II): Two mutually exclusive events with non-zero probabilities of occurrence cannot be independent.

Which of the following is correct?

If A and B are two non-mutually exclusive events such that $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B} \mid \mathrm{A})$, then

Meera visits only one of the two temples A and B in her locality. Probability that she visits temple A is $\frac{2}{5}$. If she visits temple $A, \frac{1}{3}$ is the probability that she meets her friend, whereas it is $\frac{2}{7}$ if she visits temple $B$. Meera met her friend at one of the two temples. The probability that she met her at temple B is

A die is thrown 10 times. The probability that an odd number will come up at least once is

A random variable $X$ has the following probability distribution:

If the mean of the random variable $X$ is $1 / 3$, then the variance is

If a random variable $X$ follows the binomial distribution with parameters $n=5, p$ and $P(X=2)=9 P(X=3)$, then $p$ is equal to

A bag contains $$2 n+1$$ coins. It is known that $$n$$ of these coins have head on both sides whereas, the other $$n+1$$ coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is $$\frac{31}{42}$$, then the value of $$n$$ is

Let $$A=\{x, y, z, u\}$$ and $$B=\{a, b\}$$. A function $$f: A \rightarrow B$$ is selected randomly. The probability that the function is an onto function is

If $$A$$ and $$B$$ are events, such that $$P(A)=\frac{1}{4}, P(A / B)=\frac{1}{2}$$ and $$P(B / A)=\frac{2}{3}$$, then $$P(B)$$ is

Find the mean number of heads in three tosses of a fair coin.

If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{2}, P(B)=\frac{1}{2}$$ and $$P(A \mid B)=\frac{1}{4}$$, then $$P\left(A^{\prime} \cap B^{\prime}\right)$$ is

A pandemic has been spreading all over the world. The probabilities are 0.7 that there will be a lockdown, 0.8 that the pandemic is controlled in one month if there is a lockdown and 0.3 that it is controlled in one month if there is no lockdown. The probability that the pandemic will be controlled in one month is

If $$A$$ and $$B$$ are two independent events such that $$P(\bar{A})=0.75, P(A \cup B)=0.65$$ and $$P(B)=x$$, then find the value of $$x$$.

Given that, $$A$$ and $$B$$ are two events such that $$P(B)=\frac{3}{5}, P\left(\frac{A}{B}\right)=\frac{1}{2}$$ and $$P(A \cup B)=\frac{4}{5}$$, then $$P(A)$$ is equal to

If $$A, B$$ and $$C$$ are three independent events such that $$P(A)=P(B)=P(C)=P$$, then $$P$$ (at least two of $$A, B$$ and $$C$$ occur) is equal to

Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the probability of getting a sum as 3 is

A car manufacturing factory has two plants $$X$$ and $$Y$$. Plant $$X$$ manufactures $$70 \%$$ of cars and plant $$Y$$ manufactures $$30 \%$$ of cars. $$80 \%$$ of cars at plant $$X$$ and $$90 %$$ of cars at plant $$Y$$ are rated as standard quality. A car is chosen at random and is found to be standard quality. The probability that it has come from plant $$X$$ is :

If $$P(A)=0.59, P(B)=0.30$$ and $$P(A \cap B)=0.21$$ then $$P\left(A^{\prime} \cap B^{\prime}\right)$$ is equal to

A die is thrown 10 times, the probability that an odd number will come up at least one time is

If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{3}, P(B)=\frac{1}{2}$$ and $$P(A \cap B)=\frac{1}{6}$$, then $$P\left(A^{\prime} / B\right)$$ is

Events $$E_1$$ and $$E_2$$ from a partition of the sample space $$S$$. $$A$$ is any event such that $$P\left(E_1\right)=P\left(\dot{E}_2\right)=\frac{1}{2}, P\left(E_2 / A\right)=\frac{1}{2}$$ and $$P\left(A / E_2\right)=\frac{2}{3}$$, then $$P\left(E_1 / A\right)$$ is

The probability of solving a problem by three persons $$A, B$$ and $$C$$ independently is $$\frac{1}{2}, \frac{1}{4}$$ and $$\frac{1}{3}$$ respectively. Then the probability of the problem is solved by any two of them is

If $$A, B, C$$ are three mutually exclusive and exhaustive events of an experiment such that $$P(A)=2 P(B)=3 P(C)$$, then $$P(B)$$ is equal to

Two letters are chosen from the letters of the word 'EQUATIONS'. The probability that one is vowel and the other is consonant is

A random variable '$$X$$' has the following probability distribution

Then the value of $$k$$ is

If $$A$$ and $$B$$ are two events of a sample space $$S$$ such that $$P(A)=0.2, P(B)=0.6$$ and $$P(A \mid B)=0.5$$ then $$P\left(A^{\prime} \mid B\right)=$$

If '$$X$$' has a binomial distribution with parameters $$n=6, p$$ and $$P(X=2)=12$$, $$P(X=3)=5$$ then $$P=$$

A man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set $$S=\{1,2,3,4,5,6,7\}$$ and reports that it is even. The probability that is actually even is

The probability distribution of $X$ is

$$ \begin{array}{|c|l|c|c|c|} \hline \boldsymbol{X} & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{P}(\boldsymbol{X}) & 0.3 & k & 2 k & 2 k \\ \hline \end{array} $$