AP EAPCET 2025 - 22nd May Evening Shift | Probability Question 37 | Mathematics

There are three families $F_1, F_2, F_3 . F_1$ has 2 boys and 1 girl; $F_2$ has 1 boy and 2 girls; $F_3$ has 1 boy and 1 girl. A family is randomly chosen and a child is chosen from that family randomly. If it is known that the child thus selected is a girl, then the probability that she is form $F_2$ is

$\frac{4}{9}$

$\frac{2}{9}$

$\frac{3}{7}$

$\frac{5}{7}$

AP EAPCET 2025 - 22nd May Evening Shift

MCQ (Single Correct Answer)

-0

An urn $A$ contains 4 white and 1 black ball; urn $B$ contains 3 white and 2 black balls and urn $C$ contains 2 white and 3 black balls. One ball is transferred randomly from $A$ to $B$; later one ball is transferred randomly from $B$ to $C$. Finally, if a ball is drawn randomly from $C$, then the probability that it is a black ball is

$\frac{7}{12}$

$\frac{89}{180}$

$\frac{101}{180}$

$\frac{17}{36}$

AP EAPCET 2025 - 22nd May Evening Shift

MCQ (Single Correct Answer)

-0

If the probability distribution of a discrete random variable $X$ is given by $P(X=k)=\frac{2^{-k}(3 k+1)}{2^c}, k=0,1,2, \ldots \ldots \infty$, then $P(X \leq c)=$

$\frac{\mathrm{c}}{5}$

$\frac{c}{4}$

$\frac{c+2}{5}$

$\frac{c-2}{7}$

AP EAPCET 2025 - 22nd May Evening Shift

MCQ (Single Correct Answer)

-0

In a binomial distribution, if $n=4$ and $P(X=0)=\frac{16}{81}$, then $P(X=4)=$

$\frac{1}{8}$

$\frac{1}{27}$

$\frac{1}{16}$

$\frac{1}{81}$

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