Download App

13. probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths13. probability1 Mark
MCQ
Three numbers are chosen at random, one after another with replacement, from the set $S=\{1,2,3, \ldots, 100\}$. Let $p_1$ be the probability that the maximum of chosen numbers is at least 81 and $p _2$ be the probability that the minimum of chosen numbers is at most $40$ .

($1$) The value of $\frac{625}{4} p _1$ is

($2$) The value of $\frac{125}{4} p _2$ is

Give the answer or queution ($1$) and ($2$)

  • A
    $76.35,24.70$
  • B
    $76.30,24.60$
  • C
    $76.26,24.55$
  • ✓
    $76.25,24.50$
✓

Answer

Correct option: D.
$76.25,24.50$
d
($1$) $p _1=\text { probability that maximum of chosen numbers is at least } 81$

$p _1=1-\text { probability that maximum of chosen number is at most } 80$

$p _1=1-\frac{80 \times 80 \times 80}{100 \times 100 \times 100}=1-\frac{64}{125}$

$p _1=\frac{61}{125}$

$\frac{625 p _1}{4}=\frac{625}{4} \times \frac{61}{125}=\frac{305}{4}=76.25$

the value of $\frac{625 p _1}{}$ is $76.25$

($2$) $p _2=\text { probability that minimum of chosen numbers is at most } 40$

$=1-\text { probability that minimum of chosen numbers is at least } 41$

$=1-\left(\frac{600}{100}\right)^3$

$=1-\frac{27}{125}=\frac{98}{125}$

$\therefore \frac{125}{4} p _2=\frac{125}{4} \times \frac{98}{125}=24.50$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 13. probability→13. probability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int_{}^{} {\frac{{3{x^3} - 2\sqrt x }}{x}} dx = $
→
The first derivative of the function $\left[ {{{\cos }^{ - 1}}\left( {\sin \sqrt {{{1 + x} \over 2}} } \right) + {x^x}} \right]$ with respect to $x$ at  $x = 1$  is
→
If $\tan^{-1}(\cot\theta)=2\theta,$ then $\theta=$

  1. $\pm\frac{\pi}{3}$

  2. $\pm\frac{\pi}{4}$

  3. $\pm\frac{\pi}{6}$

  4. $\text{none of these}$

→
Area of the region bounded by the curve $y=x^2$ and the line $y=4$ is
→
If $f(x)=x^2 g(x)$ and $g(x)$ is twice differentiable, then $f^{\prime \prime}(x)$ is equal to
→
The eqution of the plane through the line x + y + 3 = 0 = 2x - y + 3z + 1 and parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ is:
  1. x - 5y + 3z = 7
  2. x - 5y + 3z = -7
  3. x + 5y + 3z = 7
  4. x + 5y + 3z = -7
→
If $A \subset B$, then the value of $P(A \cap B)$ will be-
→
The corner points of the bounded feasible region are $(60,0),(120,0),(60,40),(40,20)$ and $(20,30)$. For the objective function $z=5 x+10 y \ldots$

$(i)$ Maximum value of $z$.

$(ii)$ Minimum value of $z$.

$(iii)$ Maximum value of $z$ has at

$(iv)$ Minimum value of $z$ has at

→
Bag $I$ contains $3$ red,$4$ black and $3$ white balls and Bag $II$ contains $2$ red,$5$ black and $2$ white balls. One ball is transferred from Bag $I$ to Bag $II$ and then a ball is draw from Bag $II$. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red,is.
→
Which of the following functions form Z to itself are bijections?
  1. f(x) = x3
  2. f(x) = x + 2
  3. f(x) = 2x + 1
  4. f(x) = x2 + x
→