Let $S=\left\{E_1,E_2,......,E_8\right\}$  be a sample space of a random experiment such that P(E_n) = $\frac{n}{36}$  for every n = 1, 2,……, 8. then the number of elements in the set $\left\{A\subseteq S:P\left(A\right)\ge \frac{4}{5}\right\}$ is……….

<p>P (E_n) = n/36 for n = 1, 2, 3, …., 8</p><p><math><mrow><mi>P</mi><mrow><mo> (</mo><mrow><mi>A</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mi>A</mi><mi>n</mi><mi>y</mi><mtext> </mtext><mtext> </mtext><mi>p</mi><mi>o</mi><mi>s</mi><mi>s</mi><mi>i</mi><mi>b</mi><mi>l</mi><mi>e</mi><mtext> </mtext><mtext> </mtext><mi>s</mi><mi>u</mi><mi>m</mi><mtext> </mtext><mtext> </mtext><mi>o</mi><mi>f</mi><mtext> </mtext><mtext> </mtext><mrow><mo> (</mo><mrow><mn>1</mn><mo>, </mo><mn>2</mn><mo>, </mo><mn>3</mn><mo>, </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>, </mo><mn>8</mn></mrow><mo>)</mo></mrow><mrow><mo> (</mo><mrow><mo>=</mo><mi>α</mi><mtext> </mtext><mi>s</mi><mi>a</mi><mi>y</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>3</mn><mn>6</mn></mrow></mfrac></mrow></math></p><p><br><math><mrow><mo></mo><mfrac><mrow><mi>α</mi></mrow><mrow><mn>3</mn><mn>6</mn></mrow></mfrac><mo>≥</mo><mfrac><mrow><mn>4</mn></mrow><mrow><mn>5</mn></mrow></mfrac></mrow></math></p><p><math><mrow><mo>∴</mo><mi>a</mi><mo>≥</mo><mn>2</mn><mn>9</mn></mrow></math></p><p>If one of the number from {1, 2, ….8} is left then total&nbsp;<math><mrow><mo>≥</mo></mrow></math>&nbsp;29 by 3 ways</p><p>Similarly by leaving terms more 2 or 3 we get 16 more combinations</p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;Total number of different set a possible is 16 + 3</p><p>= 19</p>

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