Three numbers are chosen at random, one after another with replac… - Vidyadip

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 $76.25$

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

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

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

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

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

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