$\begin{array}{1 1} \frac{1}{50} \\\frac{11}{50} \\ \frac{21}{50} \\ \frac{31}{50}\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Solution :

A= last card is spade

B= not spade

$P(A) = \large\frac{13}{52} =\frac {1}{4}$

$P(B) = \large\frac{39}{52} =\frac {3}{4}$

$\in $= 2 cards drawn are spade

$P \bigg( \large\frac{\in}{A} \bigg) =\bigg( \large\frac{12}{51} \bigg) \times \bigg( \large\frac{11}{50} \bigg) =\large\frac{132}{2550}$

$P \bigg( \large\frac{\in}{B} \bigg) =\bigg( \large\frac{13}{51} \bigg) \times \bigg( \large\frac{12}{50} \bigg) =\large\frac{156}{2550}$

$P\bigg(\large\frac{A}{\in}\bigg)= \large\frac{P(\Large\frac{\in}{A}) P(A)}{P(A)P(\Large\frac{\in}{A})+ P(\Large\frac{\in}{B} ) \normalsize P(B)}$

$\qquad= \large\frac{ \Large\frac{132}{2550} \times \large\frac{1}{4}}{\Large\frac{132}{2550 \times 4 } + \Large\frac{156 \times 3 }{2550 \times 4}}$

$\qquad= \large\frac{132}{132+468}$

$\qquad= \large\frac{132}{600}$

$\qquad= \large\frac{11}{50}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...