1. What are your odds of getting at least one job interview if you submit 1-10 applications?

A.
34.1%

B.
46.2%

C.
61.7%

1 answer

To calculate the odds of getting at least one job interview after submitting applications, we typically use the probability of not getting an interview for a single application and then raise that to the power of the number of applications submitted.

Assuming the chance of getting an interview with one application is a certain percentage (which is not provided in your query), the approach generally looks like this:

  1. Let \( p \) be the probability of getting an interview from one application, and thus the probability of not getting an interview from one application is \( (1 - p) \).
  2. For \( n \) applications, the probability of not getting an interview is \( (1 - p)^n \).
  3. The probability of getting at least one interview after \( n \) applications is then \( 1 - (1 - p)^n \).

If we assume a standard interview success rate common in many job markets, we might estimate \( p \) as around 0.1 (or 10%) for the sake of example:

  • For 1 application: \( 1 - (0.9)^1 = 0.1 \) (10%)
  • For 2 applications: \( 1 - (0.9)^2 = 0.19 \) (19%)
  • For 3 applications: \( 1 - (0.9)^3 = 0.271 \) (27.1%)
  • For 4 applications: \( 1 - (0.9)^4 = 0.3439 \) (34.39%)
  • For 5 applications: \( 1 - (0.9)^5 = 0.4095 \) (40.95%)
  • For 6 applications: \( 1 - (0.9)^6 = 0.4761 \) (47.61%)
  • For 7 applications: \( 1 - (0.9)^7 = 0.5443 \) (54.43%)
  • For 8 applications: \( 1 - (0.9)^8 = 0.614 \) (61.4%)
  • For 9 applications: \( 1 - (0.9)^9 = 0.6859 \) (68.59%)
  • For 10 applications: \( 1 - (0.9)^{10} = 0.7599 \) (75.99%)

Based on this example run, the chances increase with each additional application.

From the options you've provided:

A. 34.1% - This might correspond to around 4 applications. B. 46.2% - This might correspond to around 5 applications (based on a rough estimate). C. 61.7% - This is more likely aligned with around 8 applications.

Without more specific information about the individual probability \( p \), it's difficult to definitively say which answer is correct, but generally:

  • It is likely that C (61.7%) could represent a reasonable probability for a higher number of applications, like 8.