Chapter 3. Probability

Chapter 3: Probability 69Dependent, since if both, as is likely to be the case, are using the same house current, these events would bedependent.c. Drinking until your driving ability is impaired and being involved in a car crashDependent, since impaired ability to drive increases the likelihood of having a car crash3. Coin and Die, independent eventsP(tail on coin and 3 on die)= P(tail) ∗ P(3)= 1/2 ∗ 1/6= 0.5000 ∗ 0.1667= 0.08334. Letter and Digit, independent eventsP(letter K and number 9)=P(K) ∗ P(9)= 1/26 ∗ 1/10= 0.0385 ∗ 0.1000= 0.004Pregnancy Test Results. In Exercises 5-8, use the data in Table 3-1.Table 3-1. Pregnancy Test ResultsPositive Test Result(Pregnancy is indicatedNegative Test Result(Pregnancy is not indicated)Subject is pregnant 80 5 85Subject is not pregnant 3 11 14Total 83 16 99Total5. Positive Test ResultP(first test +)= 83/99= 0.8384 (one positive test is removed from data set)P(second test+)= 82/98= 0.8367P(first and second tests are both +)= P(first test +) ∗ P(second test +)=0.8384 ∗ 0.8367= 0.7016. PregnantP(tests – or is not pregnant)= P(tests –) + P(not pregnant) – P(tests – and is not pregnant)=16/99 + 14/99 – 11/99= 0.1616 + 0.1414 – 0.1111= 0.1957. PregnantP( first subject is pregnant)= 85/99= 0.8586 (one pregnant subject is removed from data set)P(second subject is pregnant)= 84/98= 0.8571P(first and second subjects are both pregnant)=P(first is pregnant) ∗ P(second is pregnant)= 0.8586 ∗ 0.8571= 0.7368. Negative Test ResultP(first one is –)= 16/99= 0.1616P(second one is –)= 15/98= 0.1531P(third is –)= 14/97)= 0.1443P(all three test –)= P(first is –) ∗ P(second is –) ∗ P(third is –)= 0.1616 ∗ 0.1531 ∗ 0.1443=0.00357