Q:

# 10 black balls and 5 white balls are placed in an urn. Two balls are then drawn in succession. What is the probability that the second ball drawn is a white ball if the second ball is drawn without replacing the first ball?a) 0.3333b) 0.2381c) 0.0952d) 0.2143e) 0.3111

Accepted Solution

A:
Answer:The probability that the second ball drawn is a white ball if the second ball is drawn without replacing the first ball is $$\frac{1}{3}$$ or 0.3333Step-by-step explanation: Probability is the greater or lesser possibility that a certain event will occur. In other words, the probability establishes a relationship between the number of favorable events and the total number of possible events. Then, the probability of any event A is defined as the ratio between the number of favorable cases (number of cases in which event A may or may not occur) and the total number of possible cases. This is called Laplace's Law:$$P(A)\frac{number of favorable cases of A}{total number of possible cases}$$Each of the results obtained when conducting an experiment is called an elementary event. The set of all elementary events obtained is called the sample space, so that every subset of the sample space is an event.The total number of possible cases is 15 (10 black balls added to the 5 white balls). As each extraction is without replacement, the events are dependent. For that, the dependent probabilities are defined first Two events are dependent on each other when the fact that one of them is verified influences the probability of the other being verified.  In other words, the probability of A happening is affected because B has happened or not.The probability of two events A and B of two successive simple experiments in a dependent compound experiment is: A and B are dependent ⇔ P (A ∩ B) = P (A) · P (B / A)                                               P (A ∩ B) = P (B) · P (B / A) As the color of the first ball that is extracted is unknown, there are two cases: that the ball is black or that the ball is white. It will be assumed first that the first ball drawn is black. Then the probability of this happening is $$\frac{10}{15}$$ since the number of black balls in the urn is 10 and the total number of cases is 15. It is now known that the second ball extracted will be white. Then the number of favorable cases will be 5 (number of white balls inside the ballot urn), but now the number of total cases is 14, because a ball was previously removed that was not replaced.  So the probability of this happening is $$\frac{5}{14}$$ So the probability that the first ball is black and the second white is:$$\frac{10}{15} *\frac{5}{14} =\frac{5}{21}$$It will now be assumed first that the first ball that is drawn is white. Then the probability of this happening is $$\frac{5}{15}$$ since the number of white balls in the urn is 5 and the total number of cases is 15. And it is known that the second ball drawn will be white. Then, the number of favorable cases will be 4 (number of white balls inside the urn, because when removing a white ball and not replacing it, its quantity will decrease), and the total number of cases is 14, same as in the previous case  So, the probability of this happening is  $$\frac{4}{14}$$ So the probability that the first ball is white and the second white is:$$\frac{5}{15} *\frac{4}{14} =\frac{2}{21}$$ If A and B are two incompatible events, that is, they cannot occur at the same time, the probability of occurrence A or of occurrence B will be the sum of the probabilities of each event occurring separately. These are events are incompatible, since I cannot, in a first extraction, extract a black and white ball at the same time. So:$$\frac{5}{21} +\frac{2}{21} =\frac{7}{21}=\frac{1}{3} =0.3333$$Finally, the probability that the second ball drawn is a white ball if the second ball is drawn without replacing the first ball is $$\frac{1}{3}$$ or 0.3333