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jeyben [28]
1 year ago
8

Choose any positive integer. Powers of two here are not very interesting, so choose something else. If the number you have chose

n is even, divide it by two. If it's odd, multiply it by three and add one. Now, with the result, repeat the same process. If the number is even divide it by two. If it's odd multiply it by three and add one. Keep repeating this process on the outcome. You will generate (*recursively*) a sequence of positive integers.
Mathematics
1 answer:
Yakvenalex [24]1 year ago
5 0

Answer:

Step-by-step explanation:

Let the integer be 6 for even and 7 for odd (say)

For 6, we divide by 2, now get 3.  Now we multiply by 3 and add 1 to get 10. Now since 10 is even divide by 5, now multiply by 3 and add 1 to get 16.  Now divide by 2 again by 2 again by 2 again by 2 till we get rid of even numbers.

The result is 1, so multiply by 3 and add 1 we get 4 now divide 2 times by 2 to get 1, thus this result now again repeats after 2 times.

Say if we select off number 3, multiply by 3 and add 1 to get 10 now divide by 5, now repeat the same process as above for 5 until we get 1 and it gets repeated every third time.

Thus whether odd or even after some processes, we get 1 and the process again and again returns to 1.

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A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of <em>X</em> is given by:

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              =\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

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