For a particular house being selected Probability =31 Prob(all the persons apply for the same house) =(31×31×31)3=91
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is
Held on 30 Apr 2005 · Verified 6 Jul 2026.
92
91
98
97
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
The probability distribution of a random variable $X$ is given below : \(\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 4k & \dfrac{30}{7}k & \dfrac{32}{7}k & \dfrac{34}{7}k & \dfrac{36}{7}k & \dfrac{38}{7}k & \dfrac{40}{7}k & 6k \\ \hline P(X) & \dfrac{2}{15} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} \\ \hline \end{array}\) If $\mathrm{E}(\mathrm{X})=\frac{263}{15}$, then $\mathrm{P}(\mathrm{X}<20)$ is equal to :
Two distinct numbers $a$ and $b$ are selected at random from $1,2,3, \ldots, 50$. The probability, that their product $a b$ is divisible by 3, is
A data consists of $20$ observations $x_1, x_2, \ldots, x_{20}$. If $\sum_{i=1}^{20}(x_i + 5)^2 = 2500$ and $\sum_{i=1}^{20}(x_i - 5)^2 = 100$, then the ratio of mean to standard deviation of this data is:
From a month of $31$ days, $3$ different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\dfrac{a}{b}$, where $a,b \in \mathbb{N}$ and $\gcd(a,b)=1$, then $a+b$ is equal to ______
Work through every JEE Main Probability & Statistics PYQ, year by year.