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8 [N2010/II/7 modified] For events 
A and 
B, it is given that 
P(A)=0.7,P(B)=0.6 and 
P(A∣B^('))=0.8. Find
(i) 
P(A nnB^(')),
(ii) 
P(A uu B),
(iii) 
P(B^(')∣A).
For a third event 
C, it is given that 
P(C)=0.5 and that 
A and 
C are independent.
(iv) Find 
P(A^(')nn C)
(v) Hence state an inequality satisfied by 
P(A^(')nn B nn C) in the form 
p <= P(A^(')nn B nn C) <= q, where 
p and 
q are constants to be determined.

88 [N20102010/II/77 modified] For events A A and B B , it is given that P(A)=0.7,P(B)=0.6 \mathrm{P}(A)=0.7, \mathrm{P}(B)=0.6 and P(AB)=0.8 P\left(A \mid B^{\prime}\right)=0.8 . Find\newline(i) P(AB) \mathrm{P}\left(A \cap B^{\prime}\right) ,\newline(ii) P(AB) \mathrm{P}(A \cup B) ,\newline(iii) P(BA) \mathrm{P}\left(B^{\prime} \mid A\right) .\newlineFor a third event C C , it is given that P(C)=0.5 \mathrm{P}(C)=0.5 and that A A and C C are independent.\newline(iv) Find B B 11\newline(v) Hence state an inequality satisfied by B B 22 in the form B B 33, where B B 44 and B B 55 are constants to be determined.

Full solution

Q. 88 [N20102010/II/77 modified] For events A A and B B , it is given that P(A)=0.7,P(B)=0.6 \mathrm{P}(A)=0.7, \mathrm{P}(B)=0.6 and P(AB)=0.8 P\left(A \mid B^{\prime}\right)=0.8 . Find\newline(i) P(AB) \mathrm{P}\left(A \cap B^{\prime}\right) ,\newline(ii) P(AB) \mathrm{P}(A \cup B) ,\newline(iii) P(BA) \mathrm{P}\left(B^{\prime} \mid A\right) .\newlineFor a third event C C , it is given that P(C)=0.5 \mathrm{P}(C)=0.5 and that A A and C C are independent.\newline(iv) Find B B 11\newline(v) Hence state an inequality satisfied by B B 22 in the form B B 33, where B B 44 and B B 55 are constants to be determined.
  1. Divide Total Amount: First, we need to divide the total amount of tape needed by the amount of tape on each roll.\newlineSo, we do 8,000cm÷2,000cm8,000 \, \text{cm} \div 2,000 \, \text{cm} per roll.
  2. Calculate Number of Rolls: Calculating that gives us 8,000÷2,000=48,000 \div 2,000 = 4 rolls.

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