Odds and Probability

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alfleshacole

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There are 5 sets of buckets labelled “A” – “E” with boxes numbered 0 to 9 in each and a 6th bucket labelled “F” with boxes numbered 0 to 20;
To qualify for the next stage of a competition, a player must be able to predict in ORDER of the buckets’ arrangement, the same box numbers as chosen by the organisers.

How many of the contestants are likely to predict in these manner:
  • Predict correctly for all 5 buckets labelled A - E and the six bucket F
  • Predict correctly for only the 5 buckets labelled A - E
  • Predict correctly for 4 of the buckets labelled A - E and the six bucket F
  • Predict correctly for only 4 of the buckets labelled A - E
  • Predict correctly for 3 of the buckets labelled A - E and the six bucket F
  • Predict correctly for only 3 of the buckets labelled A - E
  • Predict correctly for 2 of the buckets labelled A - E and the six bucket F
  • Predict correctly for only 2 of the buckets labelled A - E
 
This is a math help forum! Can you please state what have you done an where are you stuck so that we can offer you help.
If you are just looking for the solution, then you have come to the wrong site as giving you the solution would not be helpful.
I suggest that you read the posting policy for this forum.
 
Steven G said:
This is a math help forum! Can you please state what have you done an where are you stuck so that we can offer you help.
If you are just looking for the solution, then you have come to the wrong site as giving you the solution would not be helpful.
I suggest that you read the posting policy for this forum.
Click to expand...
I tried working the probability of picking at least one number from each event and multiplied all the fractions together. so the number of favourable outcome in the 5 buckets "A" - "E" is 1/10 raised to the power 5 times 1/20 from bucket "F" which has a total of 21 balls.

1/10 * 1/10 * 1/10 * 1/10 * 1/10 * 1/21 = 1 / 2,100,000
1/10 * 1/10 * 1/10 * 1/10 * 1/10 = 1 / 100,000
1/10 * 1/10 * 1/10 * 1/10 * 1/21 = 1 / 210,000
1/10 * 1/10 * 1/10 * 1/10 = 1 / 10,000
1/10 * 1/10 * 1/10 * 1/21 = 1 / 21,000
1/10 * 1/10 * 1/10 * = 1 / 1,000
1/10 * 1/10 * 1/21 = 1 / 2,100
1/10 * 1/10 = 1 / 100
 
alfleshacole said:
There are 5 sets of buckets labelled “A” – “E” with boxes numbered 0 to 9 in each and a 6th bucket labelled “F” with boxes numbered 0 to 20;
To qualify for the next stage of a competition, a player must be able to predict in ORDER of the buckets’ arrangement, the same box numbers as chosen by the organisers.

How many of the contestants are likely to predict in these manner:
  • Predict correctly for all 5 buckets labelled A - E and the six bucket F
  • Predict correctly for only the 5 buckets labelled A - E
  • Predict correctly for 4 of the buckets labelled A - E and the six bucket F
  • Predict correctly for only 4 of the buckets labelled A - E
  • Predict correctly for 3 of the buckets labelled A - E and the six bucket F
  • Predict correctly for only 3 of the buckets labelled A - E
  • Predict correctly for 2 of the buckets labelled A - E and the six bucket F
  • Predict correctly for only 2 of the buckets labelled A - E
Click to expand...
Are there are five buckets or five sets of buckets? Are you familiar with the binomial distribution?
 
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