Bertie

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UsernameBertie
Score889
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Questions 1
Answers 198

 #2
avatar+889 
+1

I think that the answer is 19C5*14C5*9C3*6C3, whatever that works out to be.

This business of dividing by 2 and 3! when you have one group of five then the other group of five is determined, or for the groups of three it doesn't matter in what order they are chosen, is wrong.

There is a difference between splitting the 19 into two groups of five and three groups of three without distinction, and splitting them into groups where each group studies a different topic.

Consider, for example, a group of four students, call them A,B,C and D, split into two twos.

4C2 = 6, so we have 6 pairs, AB, AC, AD, BC, BD, CD.

However, when the first pairing is chosen, in effect the second pairing is also chosen, so we have just three (= 6/2) possible groupings, {AB, CD}, {AC, BD} and {AD, BC}. That's the situation if it doesn't matter which pairing comes first, that is, for example, {AB, CD} is the same as {CD, AB}.

Now suppose that the students are going into two different language classes, say the first pairing study French and the second pairing study German. Now, the order of the pairings makes a difference, {AB, CD} is different from {CD, AB}.

In the first case, A and B study French and in the second case they study German, (and the other way round for C and D).

If this is the situation, the original calculation of 4C2 = 6 stands.

(Notice btw that, for the 19 student case, as far as the calculation of the final number is concerned, it doesn't matter which group is chosen first, that is, for example, 19C5*14C5*9C3*6C3 = 19C3*16C3*13C5*8C5 = ... .)

 

BobP

Bertie Sep 25, 2017
 #14
avatar+889 
+1

I'm with Melody.

 

For the 20 ball first part, the number of three ball combinations is 20C3 = 1140.

The number of ways in which we miss with all three numbers is 17C3 = 680.

The number of ways in which we have 1 hit and three misses is 3*17C2 = 408.

The number of ways in which we have 2 hits and 1 miss is 3C2*17 = 51.

The number of ways in which we have three hits is 1.

(Check that all possibilities have been covered, 680 + 408 + 51 + 1 = 1140.)

That gets us the probabilities,

zero hits 680/1140 = 0.59649 (5 dp),

1 hit 408/ 1140       = 0.35789 (5 dp),

2 hits 51/1140        = 0.04474 (5 dp),

3 hits 1/1140          = 0.00088 (5 dp).

 

An alternative method is to calculate the probabilities directly.

Using H for a hit and M for a miss, the probability of three misses MMM is (17/20)*(16/19)*(15/18) = 0.59649.

The probability of the sequence HMM is (3/20)*(17/19)*(16/18) = 0.119298, and since both of the other 1 hit sequences MHM and MMH have the same probability, the probability of 1 hit will be 3*0.119298 = 0.35789 (5 dp).

The probabilty of the two hit sequence HHM is (3/20)*(2/19)*(17/18) = 0.014912, and since both of the other two hit sequences HMH and MHH have the same probability, the probability of two hits will be 3*0.014912 = 0.04474 (5 dp).

The probability of three hits will be (3/20)*(2/19)*(1/18) = 0.00088 (5 dp).

 

That means that the probability of at least two hits (two matches) is 0.04474 + 0.00088 = 0.04562.

The probability of not winning that way will be 0.59649 + 0.35789 = 0.95438.

 

The probability of winning with the superball is 0.1 and not winning with the superball is 0.9.

 

There are four possible outcomes,

(1) not getting at least two matches and not picking the superball : probability 0.95438* 0.9 = 0.85894,

(2) getting at least two matches, but not the superball : probability                    0.04562*0.9 = 0.04106,

(3) not getting at least two matches but do pick the superball : probability        0.95438*0.1  = 0.09544,

(4) getting at least two matches and picking the superball : probability              0.04562*0.1 = 0.00456,

(the probabilities summing  to 1, as they should).

 

That means that the probability of winning is 0.04106 + 0.09544 + 0.00456 (or 1 - 0.85894) = 0.14106.

 

Tiggsy.

Bertie Jul 3, 2017