Answer:
25 rides
Step-by-step explanation:
has $28.50
15.75 + 0.50r 28.50
Subtract 15.75
0.50r 12.75
r 25.5
you can't have half rides, so 25 rides
Please help me thanks so much
2 answers:
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Answer:
Observe that the first and the second player have equal probability to get any of number. Using the principle of that symmetry, we have that
<u>Event X = 1 </u>means that the first player has got greater number than the second player, but not than the third player. So, choose any three numbers out of five of them and say that the minimal number out of these three goes to the second player, mean number to the first one and the largest to the third one. Permute remaining two numbers on remaining two people. Hence
<u>Event X = 2</u> means that the first player has got greater number than the second and the third player, but not than the fourth player. So, choose any four numbers out of five of them and say that the minimal number and the next minimal out of these four go to the second and the third player (and permute them), third number to the first one and the largest to the fourth player. Give remaining number to the last person. Hence
<u>Event X = 3</u> means that the first player has got greater number than the second, the third, and the forth player, but not than the fifth player. So, permute these five numbers as follows: give the highest to the last person, the second highest to the first, and permute remaining numbers on the remaining people. Hence
<u>Event X = 4</u> means basically the first player has won all the battles i.e, he has got the greatest number. Hence
Answer:
a) 0.984
b) 20 people
Step-by-step explanation:
a)
If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.
Now, we want to know how many samples there are with this property.
Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is
In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely
(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
where 1=carries the gene, 0=does not carry the gene.
So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.
b)
Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.