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I have this problem:

A car dealer sells 20% of cars from brand $B_1$, 30% of $B_2$ and 50% of $B_3$. It was reported that 20% of $B_1$ cars have airbags, as well as 5% and 2% for $B_2$ and $B_3$ respectively.

Given that a selected car has airbag, what is the probability of being $B_1$?

I tried to solve the problem using Bayes Theorem:

  • A: brand
  • C: has airbag

$$p(A=B_1 | C) = \frac{ p(C | A = B_1) \cdot p(A=B_1) } { p(C)}, $$

$$ p(C | A=B_1) = 0,2 \cdot 0,2 = 0,04 $$ $$ p(A=B_1) = 0,2$$

$$ p(C) = \sum_{i=1}^{n}p(A_i \cap C)\cdot p(A_i) = 0,058$$

$$ p(A = B_1 | C) = \frac{0,04 \cdot 0,2}{0,058} = 13,79\%$$

This is the way i solved, but i can't see an alternative in my test with this result. What i've done wrong?

Thanks!

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1 Answer 1

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$P(C|A=B_1)=0.2$ because you already know that the brand is $B_1$, and you don't need to multiply with $P(B_1)$ again. What you calculate is actually $P(C\cap \{A=B_1\})$.

You've made a similar mistake in the calculation of $P(C)$. It should be $$P(C)=\sum_{i=1}^3 P(C\cap \{A=B_i\})=\sum_{i=1}^3 P(C|A=B_i)P(A=B_i)$$

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  • $\begingroup$ Oh, thanks. I finally came up with the result of 0,6153. $\endgroup$
    – joann2555
    Jan 16, 2021 at 22:37

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