Colton has 16 blue marbles and 8 white ones.if he wants to place them in identical groups without any marbles left over what is
1 answer:
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Answer:
0.75 = 75% chance that only one tag is lost, provided at least one tag is lost
Step-by-step explanation:
Independent events:
If A and B are independent events, then:
Conditional probability:
Here
P(B|A) refers to the probability of event B occurring, given that event A has occurred.
is the probability of both A and B occurring together.
P(A) is the probability of event A occurring.
In this scenario:
Event A: At least one tag is missing
Event B: Only one tag is missing.
Each tag has a 40% likelihood of being lost, which is equal to 0.4.
Probability of at least one tag missing:
The events can be considered as either no tags are missing or at least one is. Their probabilities sum to 1. Thus
p is the probability that none are lost. Each tag has a 60% = 0.6 chance of not being lost, and since they are independent,
p = 0.6*0.6 = 0.36
Then
Intersection:
The intersection of at least one lost (A) and exactly one lost (B) is precisely one lost.
Then
Probability of at least one lost:
The first being lost (0.4 chance) and the second not lost (0.6 chance)
Or
The first not being lost (0.6 chance) and the second lost (0.4 chance)
So
Calculate the probability that exactly one tag is lost, given that at least one tag is lost (round to two decimal places).
0.75 = 75% likelihood that precisely one tag is lost, assuming at least one tag is lost