Team selection

Question

A team of 11 needs to be selected from 8 bolwers and 8 batsmen, a random selection is made. What is the probablility that the selected team has more bowlers than bastsmen $?$

Comments

0.5 ?

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Check my answer @Digvijaysingh Gautam.

Answer 1

This is a question of hypergeometric distribution.It is used in place of Binomial Distribution when :

a) The probability changes from trial to trial

b) Sampling is done without replacement from finite population.

So in this question ,

sample space cardinality   =   no of ways of selecting 11 players from 16 people

                                       =   ¹⁶C₁₁

                                                    =   4368

Now , but the constraint given is selected team should have more bowlers than batsmen.So we have 3 cases here :

a) No of bowlers is 6 and no of batsmen is 5 : 

Since we have to select from 8 bowlers and 8 batsmen , no of ways of doing this    =   ⁸C₆ *  ⁸C₅

b) No of bowlers is 7 and no of batsmen is 4 : 

Since we have to select from 8 bowlers and 8 batsmen , no of ways of doing this    =   ⁸C₇ *  ⁸C₄

c) No of bowlers is 8 and no of batsmen is 3 : 

Since we have to select from 8 bowlers and 8 batsmen , no of ways of doing this    =   ⁸C₈ *  ⁸C₃

So favourable outcomes  =  ⁸C₆ *  ⁸C₅ +    ⁸C₇ *  ⁸C₄ +  ⁸C₈ *  ⁸C₃

                                     =  2184

Therefore probability(more bowlers than batsmen)   = n(Favourable Outcomes)/ n(Sample Space)

                                                                          =  2184 / 4368

                                                                          =  0.5

Comments

@Habib ¹⁶C₁₁= 4368 not 4568 that's why answer should be 1/2

Think in this way there are actually only 6 cases possible

Bowlers	Batsmen
8	3
7	4
6	5
5	6
4	7
3	8

Out of these 6 ways according to question only first 3 are valid

So probability = 3/6 = 0.5

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Read the whole question properly ..

It says more bowler than batsman

and the case 4,5 and 6 of your answer is invalid..I hope u have got it now.. 

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This 6 cases are possible one. The last 3 cases are such that number of batsman are more while the first 3 are number of bowlers are more. 

probability = $\frac{Number of bowlers are more}{Number of bowlers are more+Number of batsmen are more}$ = $\frac{3}{3+3}$ = 0.5

Also in ur answer check the value of ¹⁶C₁₁. It should be 4368 not 4568. Then ur explanation will also be correct

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Corrected..