Mathematics, 29.10.2019 01:31 LucidDreamer16

Jeanne wants to start collecting coins and orders a coin collection starter kit. the kit comes with three coins chosen at random, each in a small envelope. each coin is either a nickel, a dime, or a quarter. use conditional probability to solve the following:

what is the probability that all three coins are quarters if the first two envelopes jeanne opens each contain a quarter?
what is the probability that all three coins are different if the first envelope jeanne opens contains a dime?

Answers: 1

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Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes P_B(A)

.

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

$P(D)= \frac{1}{3}$

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

Answer from: Quest

answer: $\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

$\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

which is the required equation for finding the value of x,

$\frac{x}{sin40^{\circ}}= \frac{12}{sin60^{\circ}}$

$x= 12\times \frac{sin40^{\circ}}{sin60^{\circ}}$

$x= 12\times 0.74222719896$

$x= 8.90672638762\approx 8.967$

Answer from: Quest

No he does not because he can only cover 12 square feet from 15 square feet which he will still need 3 more. this is because if 3×5=15 square feet and 15-12=3 which thats the answer

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Jeanne wants to start collecting coins and orders a coin collection starter kit. the kit comes with...