GingerAle

Solution:

\( \text {Three of the six will remain at the game until its conclusion, this leaves three (3) }\\ \text {with a 40% probability of remaining at the game until its conclusion.}\\ \text {Find the probability that two (2) or more of these three (3) will remain. }\\ \text {To find this probability, calculate the Binomial CDF for } \large \rho \small (X \geq 2, 0.40)\\ \large \binom{3}{2}*0.4^2*0.6^1 + \binom{3}{3}*0.4^3 = 0.352\\ \;\\ \text {The probability is } 35.2\%\\ \)

GA

 Many math text books give answers to selected odd number questions.

Are you imitating your text book?

GA

Is this a brain-dead moment or your standard fare?

Solution:

\(\text {Expected Value } = (1/6)*(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = \$15.17\)

GA

Sisyphus is one smart cookie too.

Solution:

The generated numbers are constrained to multiples of five, so (5) has to be used in the units position. This leaves five (5) numbers [1,2,3,4,6] to choose for the first digit, and six (6) is the only number that produces numbers greater than 500,000.  (The arrangement of the remaining numbers are irrelevant.)

The probability is 1/5

GA

This is from book VII of Euclid’s Elements.

CPhill, I thought you had your own personalized copy signed by Euclid himself.

-1 is a demerit, for butting in.

If I square my imagination, then it becomes real:

(You BUTTED in on Melody’s post!)²

You BUTTED in on Melody’s post!

Solution:

\(\text {For these types of question, use the Total Probability Theorem/Formula.}\\ \mathbb{P}(B_1) =\mathbb{P}(B_2) =\mathbb{P}(B_3) = \dfrac{1}{3}\\ \text {Probability of mint from Bag} \tiny \text {#} \\ \mathbb{P}(M|B_1) = \dfrac{1}{2}\\ \mathbb{P}(M|B_2) = \dfrac{3}{5}\\ \mathbb{P}(M|B_3) = \dfrac{2}{3}\\ \;\\ \mathbb{P}(B_1|M) = \Big (\mathbb{P}(B_1) \cdot \mathbb{P}(M|B_1)\Big)\div \Big(\mathbb{P}(B_1) \mathbb{P}(M|B1) + \mathbb{P}(B_2)\mathbb{P}(M|B_2) +\mathbb{P} P(B_3) \mathbb{P}(M|B_3)\Big)\)

\(\text {The probability that Jane chose bag 1, given that she selects a mint is equal to the probability}\\ \text {Jane chose a mint from bag #1 divided by the probability that she selected a mint.}\\\)

\(\mathbb{P}(B_1|M) = \dfrac{\Big(\dfrac{1}{3} \cdot \dfrac{1}{2}\Big) }{\Big(\dfrac{1}{3} \cdot \dfrac{1}{2}\Big) + \Big(\dfrac{1}{3} \cdot \dfrac{3}{5}\Big) +\Big( \dfrac{1}{3} \cdot \dfrac{2}{3}\Big)}\\ \;\\ \hspace{1.8cm}=\dfrac {15}{53}\\ \;\\ \hspace{1.8cm} \approx 0.28301\\ \)

GA