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# I have 20% chance to have cavity gene. If I do have the gene, there is 51% chance that I will have at least one cavity over 1 year. If I don’t have the gene, there is 19% chance that I will have at least one cavity over 1 year. Given that I have a cavity in 6 months, what’s the probability that I have at least a cavity over 1 year?

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## Réponses aux questions d'entretien

7 réponse(s)

9

You can use the Bayes theorem to calculate the posterior probability of having the cavity gene. P(gene | cavity 1) = P( cavity | gene) * P( gene) / P(cavity) = 2/5 (rounding off all the probabilities) Now, using the posterior probability, you can calculate the revised probability of having a cavity P( cavity 2 | cavity 1) = P( cavity 2 | gene) * P( gene | cavity 1) + P( cavity 2 | no gene) * P( no gene | cavity 1) = 8/25

Sarabjeet le

3

First calculate p(G | C in 6 M) = p(C in 6 M| G) p(G) /( p(C in 6 M| G)p(G) + p(C in 6 M | !G) p(!G) ) = 3/7 Now p(C in 1 Y| C in 6 M) = 0.51 * 3/7 + 0.19 * 4/7 = 2.39/7

ylchen le

2

I also got 229/700

Utilisateur anonyme le

0

The posterior probability of Gene seems to have been calculated wrong, I get 51/127!

Utilisateur anonyme le

0

Last answer has a small error in the final calculation...the answer should be 2.29/7

Utilisateur anonyme le

1

33%?

Utilisateur anonyme le

0

3/7.

Utilisateur anonyme le

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