Assume you are writing a mechanic in a video game where an item obtained from a
ID: 3221116 • Letter: A
Question
Assume you are writing a mechanic in a video game where an item obtained from a treasure chest is random. For each time the player opens the treasure chest, the item received could be a “rare” item (with probability 0.3), a “common” item (with probability 0.6), and an “empty” chest (with probability 0.1). The dropped items occur independently (i.e. the events are “memoryless”). Answer the following questions about the behavior of the dropped items:
a. What is the probability that the player obtains a common item, a rare item, and an empty chest, in that order, in three consecutive searches?
b. What is the probability that the player obtains a common item, a rare item, and an empty chest, in any order, in three consecutive searches?
c. What is the probability that the player will not obtain a common item or empty chest after five searches? d. What is the probability that the player will obtain an empty chest or common item at least once after five searches?
Explanation / Answer
A : Event of getting a rare item ; P(A) = 0.3
B : Event of getting a Common item ; P(B) = 0.6
C : Event of getting a Empty chest : P(C) = 0.1
a. Probability that the player obtains a common item, a rare item, and an empty chest, in that order, in three consecutive searches = Probability if getting a common item in first search and a rare item in the second search and an emplty chest in the thrid search = P(A and B and C)
By multiplication theorem if A,B and C are independent then
P(A and B and C) = P(A) P(B) P(C) = 0.3 x0.6x0.1 = 0.018
b. Probability that the player obtains a common item, a rare item, and an empty chest, in any order, in three consecutive searches
Number of ways getting Common item, rare item, and emplty chest in three consecutive searches in any order = 6
The different orders are : ABC,ACB,BAC,BCA,CAB,CBA
We have already computed probability of one specific order which is 0.018, The probabiity for each of other orders is also the same which is 0.018
Probability that the player obtains a common item, a rare item, and an empty chest, in any order, in three consecutive searches =6 x 0.018 = 0.108
Probability that the player obtains a common item, a rare item, and an empty chest, in any order, in three consecutive searches = 0.108
c. Probability that the player will not obtain a common item or empty chest after five searches .
It means that player gets the rare item in all the five searches
i.e P(AAAAA) = P(A)P(A)P(A)P(A)P(A) = P(A)5 = 0.35 = 0.00243
d.probability that the player will obtain an empty chest or common item at least once after five searches
Probability that the player will obtain an empty chest or common item at least once after five searches
= 1 - Probability that the player will not obtain a common item or empty chest after five searches
we have obtained Probability that the player will not obtain a common item or empty chest after five searches = 0.00243 in 'c'
Probability that the player will obtain an empty chest or common item at least once after five searches
= 1 - 0.00243 = 0.99757
Probability that the player will obtain an empty chest or common item at least once after five searches = 0.99757
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