Answers and explanations please A vase contains 100 glass balls. Exactly 20 are

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Answers and explanations please

A vase contains 100 glass balls. Exactly 20 are red and exactly 10 are green. Select two halls from the vase at random (without replacement). The remaining 70 halls in the vase are of colors other than red or green. Let R_1 be the event that the first of the two selected balls is red. Let R_2 be the event that the second of the two selected halls is red. Let G_1 be the event that the first of the two selected balls is green. Let G_2 he the event that the second of the two selected balls is green. Let A be the event that at least one of the two selected balls is red. Let B be the event that at least one of the two selected balls is green. Find the following probabilities: P(R_1) = ? P(R_1 and G_1) = ? P(R_1 and G_2) = ? P(G_2|R_1) = ? You may have noticed in the above four problems, that despite P(A|B) being defined as the quotient of P(A and B) and P(B), it's sometimes easier to compute P(A|B) than it is to compute P(A and B). For example, questions 2.1 and 2.4 are relatively easy to answer compared to 2.3. In cases like this, it "s useful to rewrite equation (1) as we did at the top to get the chain rule, P(A and B) = P(A) middot P(B|A) = P(B) middot P(A|B) This makes question 2.3 easy to answer also (allowing us to avoid any counting): P(R_1 and G_2) = P(R_1) middot P(G_2|R_1) middot P(G_2|R_1) = 20/100 middot 10/99 = 2/99 = 2.02% In 2.8-2.10 you may find the opposite to be true. I.e. you may have an easier lime finding P(A) and P(A and B) and using those two results to find P(B|A). P(R_2|G_1) = ? P(R_2 and G_1) = ? P(R_2|R_1) = ? P(A) = ? P(A and B) = ?

Explanation / Answer

2.1) probabilty of red ball=P(R1) =20/100

2.2) P(R1 and G1) =0 ; as one ball can not be both red and blue

2.3) P(R1 and G2)=P(first ball is red and second is green) =(20/100)*(10/99) =2/99

2.4)P(G2|R1) =P(if first ball is red probabilty of second green) =10/99

2.5) P(R2|G1) =P(G1 & R2)/P(G1) =(10/100)*(20/99)/(10/100) =20/99

2.6) P(R2 and G1) =(10/100)*(20/99) =2/99

2.7)P(R2|R1) =P(R1 & R2)/P(R1) =(20/100)*(19/99)/(20/100) =19/99

please revert for further clarification

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