Problem88: VAD

¡Puedes leer este texto en español! Problema88: VAD

Ej88. A company works to the production and sale of children's puzzle of 50 pieces each one. The probability that the puzzle has a print defect is 0.01 and a cut defect is 0.05.

It's possible that there are both defects independently at each piece. Calculate the average number of defectives pieces that exist at a children's puzzle.


We perform a data compilation given by problem statement:

· I ≡ 'A print defect'.
· C ≡ 'A cut defect'.

· P(I) = 0.01
· P(C) = 0.05

· D ≡ 'The piece is defect'.

· P(D) = P(I ∪ C) = P(I) + P(C) - P(I ∩ C) = 0.01 + 0.05 - 0.01·0.05 = 0.0595

· X ≡ 'Number of defectives pieces' X ~ B(50, 0.0595).

The random variable X follows a Binomial Distribution, so we apply the mean of this distribution:

· μ = n·p = 50·0.0595 = 2.975

Finally, the average number of defectives pieces is 2.975.