# Question

Once again, we will modify the light bulb in a manner similar to what was done in Exercise 3.44. Suppose we select two light bulbs to turn on when we leave the office for the weekend on Friday at 5 pm. On Monday morning at 8 am we will observe which of the light bulbs have burned out, if any. Let X be the lifetime of the first bulb and Y the lifetime of the second bulb. When we arrive at the office on Monday morning, there are four possible outcomes of the experiment:

(i) Both bulbs burned ↔ {X < 63} ∩ {Y < 63}

(ii) The first bulb burned out while the second did not ↔ {X < 63} ∩ {Y > 63},

(iii)The second bulb burned out while the first did not ↔ {X > 63} ∩ {Y < 63},

(iv)Neither bulb burned out↔ {X >63} ∩ {Y > 63}.

For each of the four cases, determine what decision should be made regarding the type of bulbs that were in the box ( i. e., L- type or S- type) and calculate the probability that the decision is wrong. As before, assume a priori probabilities of Pr (S) = 0.75 and Pr (L) = 0.25.

(i) Both bulbs burned ↔ {X < 63} ∩ {Y < 63}

(ii) The first bulb burned out while the second did not ↔ {X < 63} ∩ {Y > 63},

(iii)The second bulb burned out while the first did not ↔ {X > 63} ∩ {Y < 63},

(iv)Neither bulb burned out↔ {X >63} ∩ {Y > 63}.

For each of the four cases, determine what decision should be made regarding the type of bulbs that were in the box ( i. e., L- type or S- type) and calculate the probability that the decision is wrong. As before, assume a priori probabilities of Pr (S) = 0.75 and Pr (L) = 0.25.

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