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professor190 [17]

3 years ago

9

Question 6 (1 point)

1 answer:

Shalnov [3]3 years ago

8 0

2. Dividing exponential expression with the same base is what the left hand fractional expression is. Z^8/Z^?=z^6. Since all the bases are the same. To get the exponential power 6 on the right we subtract the denominator’s (bottom part of the fraction) exponential power (on the left) from the numerator’s (top part) exponential power.

Then, 8 - ? = 6? 2 is the only number to make the left equal the right side. This, z^8/z^2=z^6

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11111nata11111 [884]

Answer:

0.784 = 78.4% probability that there will be at least one failed graft in the next three done

Step-by-step explanation:

To solve this question, we need to understand the binomial probability distribution and conditional probability.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: No failed grafts in the first seven

Event B: At least one fail in the next three.

Intersection of events A and B:

Since the probability of a graft failling is independent of other grafts, we have that:

$P(A \cap B) = P(A)*P(B)$

So

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)*P(B)}{P(A)} = P(B)$

So we just have to find the probability of one fail in three trials.

Three trials means that $n = 3$ .

The probability is

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.4)^{0}.(0.6)^{3} = 0.216$

Then

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.216 = 0.784$

0.784 = 78.4% probability that there will be at least one failed graft in the next three done

3 0

3 years ago