i have the solution for this question , just rewrite please :

Q2/ College allow the 8 the level students two exams (attempts) for the exit exam and consider the best mark. Suppose 30% of the students fail in the first exam (FE), 20% of students fail in the second exam (SE), and 10% of students fail on both exams (FEnSE). If an 8 the level student is randomly chosen, then what is the probability that this student: (a) Fails at least one of the exam? (b) Only fails the first exam, but not the second exam? (c) Successfully passes both the exams? (d) Fails any of the exam?Q3 / A small bakery sells cookies in packages of 500 g. The cookies are handmade and the packaging is either done by the baker himself or his wife. Some customers conjecture that the wife is more generous than the baker. One customer does an experiment: he buys packages of cookies packed by the baker and his wife on 16 different days and weighs the packages. He gets the following two samples (one for the baker, onefor his wife). Weight(wife) (X) 512 530 498 540 521 528 505 523 Weight(baker)(y) 499 1500 510 495 515 503 490 511 Test whether the complaint of the customers is justified or not. Table value of t for 14 d.f at 5 % level of significance is ttab=2.145.Let A denote an event of failing at first test while i denote an event of failing at second test , we are giveni P(A) = 30 9% = 0.30 P(BJ - 20% - 0.20 A student is randomly chosen. ( 9 ) The probability that This student fails at atleast one of the tests , that is : PRALDJ = PCA) + P ( B ) - P(ADD) P ( A UB ) = 0.40 ( b) The probability that This student only fails the first test but not the second test , that is : PRAND)= PCA) - HAOR ) PCAND = 0.30 - 2.10 P( AND)= 0. 20 [ c) The probability that this student lasses both The tests , That is : P ( A A F ) = 1 - P ( A UB ) P C A DD ) = 1 0.40 P ( AND) = 0 . 30 ( d ) The probability that Student Fails any of the exam, that is : P ( A S B ) = P ( A N D ) U P CANE ) P ( L / B ) = P ( B ) _ P ( A D D + [ PEN - PCA OF ) ] PCA ($ ) = 1 30 40. 20 - 2 610 ) PA /B)= 0.30Solution: Given: n= 8 -ng = 8 nene . of days when cookies Pracked by wife. hasno . of days when cookies parked by baker Our null hypothesis is that underlying Population mean are The samp and alternative hypothesis is that population mean of weights of cookies packed by wife is greater Then population mean of weights of cooking parked by baker ine Heil = He agalast Hi:my - Me Y: sample mean of weights ( wife ) y : sample mean of weights ( laker ) X: EXi 519 625 V: E yi - 502.875 Sx. sample Varience of weights ( wife ) Sy : sample Varience of weights ( laker ) * 2 SX: 270 178.375 - (519. 625) = 168.2343 sy : Eyi V Z Sy : 252947. 625 -(502. 875) = 64.3693 [PRIMA]Sp: Pooled Varience when sample size of two grouper 5p: 5X 45x _ 168.2343+64.3593 2 SP: 116 . 29 68 51:1716 2968 = 10. 7810 Test statistic under the null hypothesis test statistic is tral: x - y SPILL + 1 - 519.625-501.875 10 - 78 40 1 1+ -519. 625-501. 875 14 , 05 = 16.75 - 310 64 10- 784- 1+ Decision Rule: Redert Ho, if teal - team at De of significance level conclusion: at 5% level of significance tipof or that = 2145 (given) Hence: Teal = 3 1064 = That = 2.145 Therefore , we reject the will hypothesis at 5 % level of Significane. therefore, we can conclude that the complaint of the cartomer is justisled that means the wife of laken is generous than the baker [PRIMA