Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e

Question

5 days ago

6

Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e

vent that the A2 right leg tag is lost. Suppose these two events are independent and P(A1)=P(A2)=0.4. Find the probability that exactly one tag is lost, given that at least one tag is lost (write it up to second decimal place).

Mathematics

1 answer:

babunello [8.4K] · Answer 1 · 2026-03-22T23:07:13+03:00

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:

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

Conditional probability:

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

Here

P(B|A) refers to the probability of event B occurring, given that event A has occurred.

$P(A \cap B)$ 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

$p + P(A) = 1$

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

$P(A) = 1 - p = 1 - 0.36 = 0.64$

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

$P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48$

Calculate the probability that exactly one tag is lost, given that at least one tag is lost (round to two decimal places).

$P(B|A) = \frac{0.48}{0.64} = 0.75$

0.75 = 75% likelihood that precisely one tag is lost, assuming at least one tag is lost