Given Information:
White balls = 6
Orange balls = 3
Required Information:
a) How many total balls are in the bag?
b) What is the probability that the 1st ball is orange?
c) What is the probability that the 2nd ball is also orange, given that the 1st ball was orange?
d) What is the probability that both the 1st and the 2nd balls are orange?
Answer:
a) Total balls = 9
b) P(1st orange) = 1/3
c) P(2nd orange | 1st orange) = 1/4
d) P(1st orange and 2nd orange) = 1/12
Step-by-step explanation:
a)
There are 6 white balls and 3 orange balls,
Total balls = 6 + 3
Total balls = 9
b)
The probability that the first ball is orange is
P(1st orange) = No. of orange balls/Total balls
P(1st orange) = 3/9
P(1st orange) = 1/3
c)
The probability that the second ball is also orange, given that the first ball was orange is
If we assume that the 1st ball was orange then there are 2 orange balls left and total 8 balls left (since replacement is not allowed)
P(2nd orange | 1st orange) = No. of orange balls/Total balls
P(2nd orange | 1st orange) = 2/8
P(2nd orange | 1st orange) = 1/4
d)
The probability that both 1st and 2nd balls are orange is
P(1st orange and 2nd orange) = P(1st orange)*P(2nd orange | 1st orange)
P(1st orange and 2nd orange) = 1/3*1/4
P(1st orange and 2nd orange) = 1/12
![\bf \qquad \qquad \textit{Future Value of an ordinary annuity} \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right] \\\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7BFuture%20Value%20of%20an%20ordinary%20annuity%7D%0A%5C%5C%5C%5C%0AA%3Dpymnt%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%20%5Cright%29%5E%7Bnt%7D-1%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C)
![\bf \begin{cases} A= \begin{array}{llll} \textit{original amount}\\ \textit{already compounded} \end{array} & \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &400\\ r=rate\to 10\%\to \frac{10}{100}\to &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, so once} \end{array}\to &1\\ t=years\to &14 \end{cases} \\\\\\ A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right] ](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Boriginal%20amount%7D%5C%5C%0A%5Ctextit%7Balready%20compounded%7D%0A%5Cend%7Barray%7D%20%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%0A%5Cend%7Barray%7D%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26400%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.1%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20so%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0A%0At%3Dyears%5Cto%20%2614%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D400%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.1%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%2014%7D-1%7D%7B%5Cfrac%7B0.1%7D%7B1%7D%7D%20%5Cright%5D%0A)