JEE Main 2022 — Mathematics Probability & Statistics

Given,

$$P(A)\geq \frac{4}{5}$$P(A)≥54​

$A$A is subset of $S$S hence

$A$A can have elements;

type $1$1:${}$

type $2$2: ${{E}_{1}},{{E}_{2}},\ldots \ldots {{E}_{8}}$E1​,E2​,……E8​

type $3$3: ${{E}_{1},{E}_{2}},{{E}_{1},{E}_{3}}\ldots \ldots {{E}_{1},{E}_{8}}$E1​,E2​,E1​,E3​……E1​,E8​

$$\vdots$$⋮

$$\vdots$$⋮

type $6$6: ${{E}_{1},{E}_{2},\ldots \ldots .{E}_{5}},\ldots \ldots {{E}_{4},{E}_{5},{E}_{6},{E}_{7},{E}_{8}}$E1​,E2​,…….E5​,……E4​,E5​,E6​,E7​,E8​

type $7$7: ${{E}_{1},{E}_{2},\ldots \ldots .{E}_{6}},\ldots \ldots {{E}_{3},{E}_{4},\ldots \ldots \ldots .{E}_{8}}$E1​,E2​,…….E6​,……E3​,E4​,……….E8​

type $8$8: ${{E}_{1},{E}_{2},\ldots \ldots .{E}_{7}}{{E}_{2},{E}_{3},\ldots \ldots .{E}_{8}}$E1​,E2​,…….E7​E2​,E3​,…….E8​

type $9$9: ${{E}_{1},{E}_{2},\ldots \ldots {E}_{8}}$E1​,E2​,……E8​

As $P(A)\geq \frac{4}{5}$P(A)≥54​;

Note: Type $1$1 to Type $4$4 elements can not be in set $A$A as maximum probability of type $4$4 elements.

${{E}_{5},{E}_{6},{E}_{7},{E}_{8}}$E5​,E6​,E7​,E8​ is $\frac{5}{36}+\frac{6}{36}+\frac{7}{36}+\frac{8}{36}=\frac{13}{18}<\frac{4}{5}$365​+366​+367​+368​=1813​<54​

Now for Type $5$5 acceptable elements let's call probability as ${P}_{5}$P5​

$${P}_{5}=\frac{{n}_{1}+{n}_{2}+{n}_{3}+{n}_{4}+{n}_{5}}{36}\geq \frac{4}{5}$$P5​=36n1​+n2​+n3​+n4​+n5​​≥54​

$$\Rightarrow {n}_{1}+{n}_{2}+{n}_{3}+{n}_{4}+{n}_{3}\geq 28.8$$⇒n1​+n2​+n3​+n4​+n3​≥28.8

Hence, $2$2 possible ways ${{E}_{5},{E}_{6},{E}_{7},{E}_{8},{E}_{3}\mathrm{or}{E}_{4}}$E5​,E6​,E7​,E8​,E3​orE4​

$${P}_{6}={n}_{1}+{n}_{1}+{n}_{3}+{n}_{4}+{n}_{3}+{n}_{6}\geq 28.8$$P6​=n1​+n1​+n3​+n4​+n3​+n6​≥28.8

$\Rightarrow 9$⇒9 possible ways

$${P}_{7}\Rightarrow {n}_{1}+{n}_{1}+\ldots \ldots \ldots +{n}_{7}\geq 28.8$$P7​⇒n1​+n1​+………+n7​≥28.8

$\Rightarrow 7$⇒7 possible ways

$${P}_{8}\Rightarrow {n}_{1}+{n}_{1}+\ldots \ldots +{n}_{n}\geq 28.8$$P8​⇒n1​+n1​+……+nn​≥28.8

$\Rightarrow 1$⇒1 possible way

Total $=19$=19