A group major medical policy is written with a $1,000 deductible, 80/20 coinsurance, and an out-of-pocket maximum of $3,000. The total cost of the procedure is $5,000. How much does the subscriber have to pay towards the $5,000 total?

Rationale

The subscriber has to pay $1,800 towards the $5,000 total.

To determine the subscriber's payment, we first apply the deductible of $1,000. After this, the remaining amount of $4,000 is subject to the 80/20 coinsurance, where the subscriber pays 20% of that amount. This results in an additional payment of $800, leading to a total out-of-pocket cost of $1,800.

A ($5,000) YOUR ANSWER

This choice implies that the subscriber would pay the entire cost of the procedure, which is incorrect. The plan includes a deductible, coinsurance, and an out-of-pocket maximum, all of which reduce the subscriber's total payment.

B ($3,000) YOUR ANSWER

While this option suggests a payment amount close to the out-of-pocket maximum, it does not account for the deductible and the subsequent coinsurance properly. The calculation reveals a lower total payment than the maximum due to the structure of the insurance plan.

D ($1,000) YOUR ANSWER

This choice only reflects the deductible amount and fails to include the coinsurance payment that follows. After the deductible is met, additional costs remain, which must be factored into the total subscriber payment.

Conclusion

In summary, the correct total amount the subscriber must pay towards the $5,000 procedure cost is $1,800, which includes the $1,000 deductible and the $800 from the 20% coinsurance on the remaining amount. This calculation underscores the importance of understanding how deductibles and coinsurance interact within a health insurance policy.

A group major medical policy is written with a $1,000 deductible, 80/20 coinsurance, and an out-of-pocket maximum of $3,000. The total cost of the procedure is $5,000. How much does the subscriber have to pay towards the $5,000 total?

The subscriber has to pay $1,800 towards the $5,000 total.

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