Probability of choosing
i) medium type pencil is = 0.593,
ii) heavy type pencil = 0.135
iii) not very fine = 0.966
iv) neither heavy or medium = 1.458
In the question it is given that there are :
2 packs = extra fine pencil (3H)
14 packs = fine pencil (2H)
35 packs = medium pencil (HB)
8 packs = heavy pencil (2B)
Total there are 59 packs in that wholesale stocks
Let , A = extra fine pencil (3H)
B = fine pencil (2H)
C = medium pencil (HB)
D = heavy pencil (2B)
Now we have to find the probability,
i) P(C) = 35/59 =0.593
ii) P(D) = 8/59 = 0.135
iii) P(A) = 1-(2/59) =0.966
iv) P(D ∩ C) = 1- P(D U C )
= 1 – [P(D) + P(C)] (As they are disjoint events)
= 1- [ (8/59) – (35/59)]
= 1 + (27/59)
= 1.458
Hence, probability of choosing pencil of type
i) medium = 0.593,
ii) heavy = 0.135
iii) not very fine = 0.966
iv) neither heavy or medium = 1.458
Learn more about probability here : https://brainly.com/question/25870256
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