Question

3. Three scientists are trying to meet in NWA for a research experiment Scientist 1 is arriving on a flight from China, Scientist 2 is arriving on a flight from Singapore, and Scientist 3 is arriving on a flight from UK. Historical data suggests the flights from China, Singapore, and UK are "on time" 83%,96%, and 90% of the time, respectively. Historical data also suggests the three flights are independent with respect to on time behavior. a) Define the sample space for this random experiment, indicating which scientists are present for the meeting in each outcome in the sample space b) Compute the probability for each of the outcomes in the sample space 4. If two cards are drawn from a regular pack of 52 playing cards, what is the probability: a) The first card drawn is an ace? b) That the second card drawn is an ace, given that the first card drawn (which was not replaced in the deck) was an ace? c) That the second card drawn is an ace, given that the first card (which was not replaced in the deck) was NOT an ace?

Answer 1

let event A,B and C represent that scientist I ,II and III are on time in meeting

and Ac ; Bc and Cc represent that scientist I ,II and III are not on time in meeting

hence a) sample space ={ AcBcCc , ABcCc , AcBCc , AcBcC , ABCc ,ABcC ,AcBC ,ABC}

b) P(AcBcCc)=(1-0.83)*(1-0.96)*(1-0.90)=0.00068

P(ABcCc)=(0.83)*(1-0.96)*(1-0.90)=0.00332

P(AcBCc)=(1-0.83)*(0.96)*(1-0.90)=0.01632

P(AcBcC)=(1-0.83)*(1-0.96)*(0.90)=0.00612

P(ABCc)=0.83)*(0.96)*(1-0.90)=0.07968

P(ABcC)=(0.83)*(1-0.96)*(0.90)=0.02988

P(AcBC)=(1-0.83)*(0.96)*(0.90)=0.14688

P(ABC)=(0.83)*(0.96)*(0.90)=0.71712

4)

a)P(first card is ace)=4/52 =1/13 (as there are 4 ace out of 52 cards)

b)P(second ace given first ace) =3/51 =1/17 (as there remains 3 ace out of 51 cards if first card is ace)

c)

P(second ace given first was not ace)=4/51 (as there remains 4 ace out of 51 cards if first card is not an ace)