# Recent questions tagged probability

1 vote
1
If $A$ and $B$ are two related events, and $P(A \mid B)$ represents the conditional probability, Bayes’ theorem states that $P(A\mid B) = \dfrac{P(A)}{P(B)} P(B\mid A)$ $P(A\mid B) = P(A) P(B) P(B\mid A)$ $P(A\mid B) = \dfrac{P(A)}{P(B)}$ $P(A\mid B) = P(A)+P(B)$
2
If $P$ is risk probability, $L$ is loss, then Risk Exposure $(RE)$ is computed as. $RE = P/L$ $RE = P + L$ $RE = P \ast L$ $RE = 2 \ast P \ast L$
1 vote
3
If a random coin is tossed $11$ times, then what is the probability that for $7$th toss head appears exactly $4$ times? $5/32$ $15/128$ $35/128$ None of the options
1 vote
4
If $X, Y$ and $Z$ are three exhaustive and mutually exclusive events related with any experiment and the $P\left(X \right)=0.5P\left(Y \right)$ and $P\left(Z \right)$ = $0.3P\left(Y \right)$. Then $P\left(Y \right)$ = ___________ . $0.54$ $0.66$ $0.33$ $0.44$
5
A box contains $10$ screws, $3$ of which are defective. Two screws are drawn at random with replacement. The probability that none of two screws is defective will be $100\%$ $50\%$ $49\%$ None of these.
1 vote
6
The probability that top and bottom cards of a randomly shuffled deck are both aces is: $4/52\times 4/52$ $4/52\times 3/52$ $4/52\times 3/51$ $4/52\times 4/51$
7
Let $\mathcal{R}$ be the set of all binary relations on the set $\{1,2,3\}$. Suppose a relation is chosen from $\mathcal{R}$ at random. The probability that the chosen relation is reflexive (round off to $3$ decimal places) is ______.
8
In a certain year, there were exactly four Fridays and exactly four Mondays in January. On what day of the week did the $20^{th}$ of the January fall that year (recall that January has $31$ days)? Sunday Monday Wednesday Friday None of the others
9
A lottery chooses four random winners. What is the probability that at least three of them are born on the same day of the week? Assume that the pool of candidates is so large that each winner is equally likely to be born on any of the seven days of the week independent of the other winners. ... $\dfrac{48}{2401} \\$ $\dfrac{105}{2401} \\$ $\dfrac{175}{2401} \\$ $\dfrac{294}{2401}$
10
Fix $n\geq 4.$ Suppose there is a particle that moves randomly on the number line, but never leaves the set $\{1,2,\dots,n\}.$ Let the initial probability distribution of the particle be denoted by $\overrightarrow{\pi}.$ In the first step, if the particle is at position $i,$ it moves to one ... $i\neq 1$ $\overrightarrow{\pi}(n) = 1$ and $\overrightarrow{\pi}(i) = 0$ for $i\neq n$
11
Two balls are drawn uniformly at random without replacement from a set of five balls numbered $1,2,3,4,5.$ What is the expected value of the larger number on the balls drawn? $2.5$ $3$ $3.5$ $4$ None of the above
1 vote
12
For the distributions given below : Which of the following is correct for the above distributions? Standard deviation of $A$ is significantly lower than standard deviation of $B$ Standard deviation of $A$ is slightly lower than standard deviation of $B$ Standard deviation of $A$ is same as standard deviation of $B$ Standard deviation of $A$ is significantly higher than standard deviation of $B$
1 vote
13
So, I have read the birthday paradox problem, and now I came across below question: Assuming the following: there are no leap years, all years have $n = 365$ days and that people's birthdays are uniformly distributed across the $n$ days of the year. (i) How many people must be there in a ... $n=23$, this works out to be 0.53 and Yes it seems to me I am done. Please correct me If I am wrong.
1 vote
14
Each day, you independently decide, with probability p, to flip a fair coin. Otherwise, you do nothing. (a) What is the probability of getting exactly 10 Heads in the first 20 days? (b) What is the probability of getting 10 Heads before 5 Tails?
1 vote
15
A permutation of $1,2, \dots, n$ is chosen at random. Then the probability that the numbers $1$ and $2$ appear as neighbour equals $\frac{1}{n}$ $\frac{2}{n}$ $\frac{1}{n-1}$ $\frac{1}{n-2}$
1 vote
16
Two coins are tossed independently where $P$(head occurs when coin $i$ is tossed) $=p_i, \: i=1,2$. Given that at least one head has occurred, the probability that coins produced different outcomes is $\frac{2p_1p_2}{p_1+p_2-2p_1p_2}$ $\frac{p_1+p_2-2p_1p_2}{p_1+p_2-p_1p_2}$ $\frac{2}{3}$ none of the above
1 vote
17
Let $A_1,A_2,A_3, \dots , A_n$ be $n$ independent events such that $P(A_i) = \frac{1}{i+1}$ for $i=1,2,3, \dots , n$. The probability that none of $A_1, A_2, A_3, \dots , A_n$ occurs is $\frac{n}{n+1}$ $\frac{1}{n+1}$ $\frac{n-1}{n+1}$ none of these
1 vote
18
A determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability of choosing a non-zero determinant is $\frac{3}{16}$ $\frac{3}{8}$ $\frac{1}{4}$ none of these
1 vote
19
A basket contains some white and blue marbles. Two marbles are drawn randomly from the basket without replacement. The probability of selecting first a white and then a blue marble is $0.2$. The probability of selecting a white marble in the first draw is $0.5$. What is the probability of ... a blue marble in the second draw, given that the first marble drawn was white? $0.1$ $0.4$ $0.5$ $0.2$
20
If $P$ is an integer from $1$ to $50$, what is the probability that $P(P+1)$ is divisible by $4$? $0.25$ $0.50$ $0.48$ none of these
1 vote
21
A die is thrown thrice. If the first throw is a $4$ then the probability of getting $15$ as the sum of three throws is $\frac{1}{108}$ $\frac{1}{6}$ $\frac{1}{18}$ none of these
Suppose you alternate between throwing a normal six-sided fair die and tossing a fair coin. You start by throwing the die. What is the probability that you will see a $5$ on the die before you see tails on the coin? $\frac{1}{12}$ $\frac{1}{6}$ $\frac{2}{9}$ $\frac{2}{7}$
If it is assumed that all $\binom{52}{5}$ ... I'm getting answer 0.095 but in the book answer is given 0.0475 where am I going wrong?
Compute the probability that if 10 married couples are seated at random at a round table, then no wife sits next to her husband 1 wife sits next to her husband. pick one of the 10 couples=$\binom{10}{1}$. These couples can interchange their position such that they sit next ... one couple sits together=$\frac{N}{19!}$ so probability that no couple sits together=$1-\frac{N}{19!}$ is this correct?