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zysi [14]
3 years ago
9

50 question true-false final exam (two-points each question) and you never paid attention in the class (or knew anything about t

he topic on your own), what is probability you will receive a grade of 56 or less, in which case you will fail the class? What is the probability you will get at least 30 questions right, which will give you a passing grade in the class?
Mathematics
1 answer:
DanielleElmas [232]3 years ago
3 0

Title:

<h2>Check the explanation.</h2>

Step-by-step explanation:

As all the questions of the exam carry 2 points, at least \frac{56}{2} = 28 questions need to be correct to get 56 marks.

In order to get a maximum of 56 marks, at most 28 answers needs to be correct.

Every question has two option. Hence, the probability of getting a particular question right is \frac{1}{2}.

The probability that exactly n questions will be right is ^{50}C_n (\frac{1}{2} )^n (\frac{1}{2} )^{50 - n} = ^{50}C_n (\frac{1}{2} )^{50}.

The probability that at most 28 questions will be right is ∑^{50}C_n (\frac{1}{2} )^{50}, 0\leq n\leq 28.

Similarly, the probability that at least 30 questions will be right is ∑^{50}C_n (\frac{1}{2} )^{50}, 30\leq n \leq 50.

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Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

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1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

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(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

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