### Problem Statement:#### Astrobiology and Probability---Suppose that there is a 0.9% chance a planet will develop life, a 2.3% chance that the life will evolve into complex organisms, a 7% chance that complex organisms develop into advanced civilizations, and an X% chance that the advanced civilizations become intergalactic.If there are 2,131,984 planets, and we observe only 1 intergalactic civilization, what is the expected value of X?---**Note:**- X is in percentage form (e.g., 30.51 not 0.3051).- You can round to two decimal places, but do not round until your final answer!---### Explanation:Consider each step as a sequential probability process:1. **Probability that a planet develops life:** $ P(\text{Life}) = 0.009 $ 2. **Probability that life evolves into complex organisms, given that life has already developed:** $ P(\text{Complex Organisms}|\text{Life}) = 0.023 $ 3. **Probability that complex organisms develop into advanced civilizations, given that complex organisms have developed:** $ P(\text{Advanced Civilizations}|\text{Complex Organisms}) = 0.07 $ 4. **Probability that advanced civilizations become intergalactic, given that advanced civilizations exist (this is X, which we need to find):** $ P(\text{Intergalactic}|\text{Advanced Civilizations}) = \frac{X}{100} $ ### Formulating the Combined Probability: The combined probability for one planet to reach the intergalactic civilization stage is:\[ P(\text{Intergalactic}) = P(\text{Life}) \times P(\text{Complex Organisms}|\text{Life}) \times P(\text{Advanced Civilizations}|\text{Complex Organisms}) \times P(\text{Intergalactic}|\text{Advanced Civilizations}) \]### Setting Up the Equation: With 2,131,984 planets and 1 observed intergalactic civilization: \[ 1 = 2,131,984 \times 0.009 \times 0.023 \times 0.07 \times \frac{X}{100} \]\[ 1 =

Given Probability of a planet can develop life=0.9% Probability of a planet can evolve into…

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A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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8.7 q.13 Please fill in the blank box with the correct answer, thank you.
PART C. 1. James is an operator of a machine that uses 16 batteries. Because it has so many, it does not shut off until half of the batteries are dead. a. All batteries are independent and have uniformly distributed lifetimes on the interval [ 0, 1 2] years. Find the probability that the machine shuts before 1 years of work? b. What is the expected time when the 5th and the 8th battery die?
In the game of roulette, a player can place a $10 bet on the number 13 and have a probability of winning. If the metal ball lands on 13, the player gets to keep the 38 $10 paid to play the game and the player is awarded an additional $350. Otherwise, the player is awarded nothing and the casino takes the player's $10. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose. The expected value is $ (Round to the nearest cent as needed.)
8.
In the game of roulette, a player can place a $9 bet on the number 29 and have a probability of winning. If the metal ball lands on 29, the player gets to keep the 38 1 $9 paid to play the game and the player is awarded an additional $315. Otherwise, the player is awarded nothing and the casino takes the player's $9, Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose. The expected value is $.5 (Round to the nearest cent as needed.)
In the game of roulette, a player can place a $7 bet on the number 32 and have a 138 probability of winning. If the metal ball lands on 32, the player gets to keep the $7 paid to play the game and the player is awarded an additional $245.Otherwise, the player is awarded nothing and the casino takes the player's $7. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose.
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Example 2.4.6 A series of tests on a certain type of electrical relay, have revealed that in approximately 5% of the trials, the relay fails to operate under the specified conditions. What is the probability that in ten trials made under these conditions the relay will fail to operate one or more times?
QUESTION 3 A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 10%. If the inspector of the retailer randomly picks 18 items from a shipment, the probability that there will be at least one defective item among these 18 items is O 0.5497 O 0.1501 O 0.4503 O 0.8499 lick Save and Submit to save and submit. Click Save All Answers to save all answers.
Question 4 Suppose that in a population of twins, males (M) and females (F) are equally likely to occur and that the probability that twins are identical is 0. If twins are not identical, their genders are independent (if they are identical, their genders are the same). (i) Show that, ignoring birth order, P(MM) = P(FF) = (1+0)/4 and P(MF) = (1–0)/2. (ii) Suppose that n twins are sampled. It is found that n₁ are MM, n2 are FF, and në are MF, but it is not known which twins are identical. Find the maximum likelihood estimator of 0.
In the game of roulette, a player can place a $6 bet on the number 11 and have a ag probability of winning. If the metal ball lands on 11, the player gets to keep the $6 paid to play the game and the player is awarded an additional $210. Otherwise, the player is awarded nothing and the casino takes the player's $6. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose. The expected value is $ (Round to the nearest cent as needed.)
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