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If p(x) = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGabaWdaeaafaqabeGabaaabaWdbmaalaaapaqaa8qacaWG4baa % paqaa8qacaaIXaGaaGynaaaacaGG7aGaamiEaiabg2da9iaaigdaca % GGSaGaaGOmaiaacYcacaaIZaGaaiilaiaaisdacaGGSaGaaGynaaWd % aeaapeGaaGimaiaacUdacaWGLbGaamiBaiaadohacaWGLbGaam4Dai % aadIgacaWGLbGaamOCaiaadwgaaaGaai4jaaGaay5EaaGaaiiOaiaa % dshacaWGObGaamyzaiaacckacaWGWbGaamOCaiaad+gacaWGIbGaam % yyaiaadkgacaWGPbGaamiBaiaadMgacaWG0bGaamyEaiaacckacaWG % qbGaaiiOamaacmaapaqaa8qadaWcaaWdaeaapeGaaGymaaWdaeaape % GaaGOmaaaacqGH8aapcaWGybGaeyipaWZaaSaaa8aabaWdbiaaiwda % a8aabaWdbiaaikdaaaaacaGL7bGaayzFaaaaaa!6956! \left\{ {\begin{array}{*{20}{c}} {\frac{x}{{15}};x = 1,2,3,4,5}\\ {0;elsewhere} \end{array}'} \right.\;the\;probability\;P\;\left\{ {\frac{1}{2} < X < \frac{5}{2}} \right\}\)\(\left\{\begin{matrix} \rm\frac{x}{15};&\rm x = 1,2,3,4,5 \\ 0;&\rm \text{elsewhere} \end{matrix}\right.\), the probability \(\rm P\left\{ \frac{1}{2} < X < \frac{5}{2}\right\}\) is equal to:

If p(x) = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaGabaWdaeaafaqabeGabaaabaWdbmaalaaapaqaa8qacaWG4baa % paqaa8qacaaIXaGaaGynaaaacaGG7aGaamiEaiabg2da9iaaigdaca % GGSaGaaGOmaiaacYcacaaIZaGaaiilaiaaisdacaGGSaGaaGynaaWd % aeaapeGaaGimaiaacUdacaWGLbGaamiBaiaadohacaWGLbGaam4Dai % aadIgacaWGLbGaamOCaiaadwgaaaGaai4jaaGaay5EaaGaaiiOaiaa % dshacaWGObGaamyzaiaacckacaWGWbGaamOCaiaad+gacaWGIbGaam % yyaiaadkgacaWGPbGaamiBaiaadMgacaWG0bGaamyEaiaacckacaWG % qbGaaiiOamaacmaapaqaa8qadaWcaaWdaeaapeGaaGymaaWdaeaape % GaaGOmaaaacqGH8aapcaWGybGaeyipaWZaaSaaa8aabaWdbiaaiwda % a8aabaWdbiaaikdaaaaacaGL7bGaayzFaaaaaa!6956! \left\{ {\begin{array}{*{20}{c}} {\frac{x}{{15}};x = 1,2,3,4,5}\\ {0;elsewhere} \end{array}'} \right.\;the\;probability\;P\;\left\{ {\frac{1}{2} < X < \frac{5}{2}} \right\}\)\(\left\{\begin{matrix} \rm\frac{x}{15};&\rm x = 1,2,3,4,5 \\ 0;&\rm \text{elsewhere} \end{matrix}\right.\), the probability \(\rm P\left\{ \frac{1}{2} < X < \frac{5}{2}\right\}\) is equal to: Correct Answer <span class="math-tex">\(\frac{1}{5}\)</span>

 

Given

P(x) = {x/15 ; x = 1, 2, 3, ----5

           {0 ;        otherwise

Calculation

Probability P{1/2 < X < 5/2} that means, X = 1, 2 are only applicable

∴P(x = 1) = 1/15

∴ P(x = 2) = 2/15

⇒ P(x ) = p(x = 1) + p(x = 2)

⇒ p(x) = 1/15 + 2/15

∴ The probability of p(x) is 3/15 or 1/5

 

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