A bag contains $100$ balls numbered from $1$ to $100$ . Four balls are removed at random without replacement. What is the probability that the number on the last ball is greater than the number on first ball?

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Solution

$i f f i r s t b a l l i s 1 t h e n n u m b e r o f w a y s o f s e l e c t i n g s e c o n d b a l l = 99 i f f i r s t b a l l i s 2 t h e n . . . . . . . . . . . . . . . . . . . . . . . . . . . .98 a n d s o o n H e n c e p r o b a b i l i t y = P (1) P (> 1) + P (2) P (> 2) + . . . . . . . . . . . . = (\frac{1}{100}) t i m e s (\frac{99}{90} + (\frac{1}{100})) \times (\frac{98}{90}) + . . . . . . . . + (\frac{1}{100}) \times (\frac{1}{90}) = (\frac{99 + 98 + . . . . + 1}{100 \times 99}) = (\frac{(\frac{99 \times (99 + 1)}{2})}{100 \times 99}) = (\frac{1}{2})$

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A bag contains 100 balls numbered from 1 to 100. Four balls are removed at random without replacement. What is the probability that the number on the last ball is greater than the number on first ball?

iffirstballis1thennumberofwaysofselectingsecondball=99iffirstballis2then............................98andsoonHenceprobability=P(1)P(>1)+P(2)P(>2)+............=(1100)times(9990+(1100))×(9890)+........+(1100)×(190)=(99+98+....+1100×99)=((99×(99+1)2)100×99)=(12)

A bag contains $100$ balls numbered from $1$ to $100$ . Four balls are removed at random without replacement. What is the probability that the number on the last ball is greater than the number on first ball?