edited by
2,312 views

2 Answers

5 votes
5 votes

5 balls we have.

2 balls can be drawn in 10 ways 

Ball1, Ball2

Ball1, Ball3

Ball1, Ball4

Ball1, Ball5

Ball2, Ball3

Ball2, Ball4

Ball2, Ball5

Ball3, Ball4

Ball3, Ball5

Ball4, Ball5

Let x be the larger number on the balls drawn

x P(x)
1 0
2 1
3 2
4 3
5 4

 

We have to calculate $\sum_{1}^{5} x P(x)$

$1(0) + 2(\frac{1}{10})+ 3(\frac{2}{10})+ 4(\frac{3}{10})+ 5(\frac{4}{10}) = 4$

3 votes
3 votes

Total ways of drawing the balls = 5C2 = 10

Note: We consider (x, y) and (y, x) as same, as order doesn't matter here.

Case 1: suppose 1 is present in the balls drawn, then the greater number could be 2, 3, 4, 5

Case 2: suppose 2 is present in the balls drawn, then the greater number could be 3, 4, 5

Case 3: suppose 3 is present in the balls drawn, then the greater number could be 4, 5

Case 3: suppose 4 is present in the balls drawn, then the greater number could be 5 only

Case 5: if 5 is present in one of the balls, this case is already considered in above parts.

Now, expected value of larger number in all the possible draws = (2 + 3 + 4 + 5 + 3 + 4 + 5 + 4 + 5 + 5) / 10 = 40 / 10 = 4

Answer:

Related questions

1 votes
1 votes
2 answers
1
admin asked Feb 10, 2020
2,138 views
In a certain year, there were exactly four Fridays and exactly four Mondays in January. On what day of the week did the $20^{th}$ of the January fall that year (recall th...