Question

im a little confused on how to work this problem. please help explain how and why. thank you.

Exercise 12. Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card and B the analogous event for MasterCard. Suppose that P(A) = 0.6 and P(B) = 0.4. a. Could it be the case that P(ANB) = 0.5? Why or why not? b. From now on, suppose that P(ANB) = 0.3. What is the probability that the selected student has at least one of these two types of cards? c. Describe, in terms A and B, the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event. d. Calculate the probability that the selected student has exactly one of the two types of cards.

Answer 1

Goodluck haja No Selected two exactly student cards- one of the has AUB PCAUB) = PCA) + PCB) - PCACB)

GoodLuck Page No. Date A selected student has a Visa Card. PCA) = 0.6 B: Selected student P(B) = 0.4 has a Master Card. .a. liven that PC AnB) = 0.5 we have, - also AnBCA Ane ce. A Since we have - AnB is subset of PCAMB) SPCA). thus here ons <0,6 subset of B. sp(B) Also AnB is : PCANB) but here PCAOB)=ois > which is not PCB) = 0,4 possible PRANB) = o.s is not possible

PRAOB 0.3 P one et selected these student two types has of at least land - P (student has none of the cards) have I P (student does not master card). X - P(student does not to have visa rard = 1 - [ C1-004) *(1-0.6)] 1- 0.68 0.47 - 1 - 0.24 = 0.76 c. Selected student has a Visa Card - A selected student does not has master Card-B . Required event is ANBI PCAOB) = P(A)-PCANB) 0.6 - 0.3 0.3