<span>The urn contains 2 purple balls and 4 white balls. The player pay $4 for start the game and get $1.5 for every ball drawn until one purple ball is drawn. The maximal revenue would be $7.5 when 4 white balls and 1 purple balls are drawn.
If the purple ball is p and white ball is w, t</span>he possible sample space of drawings are {p, wp, wwp, wwwp, wwwwp}
<span>1. Write down the probability distribution for the player earning
The player earning </span>for each event depends on the number of balls drawn subtracted the ticket price.<span>
p= 2/6
The player earnings would be: 1*$1.5 -$4= - $2.5
wp= (4*2)/(6*5) = 4/15
</span>The player earnings would be: 2*1.5- $4= - $1
wwp= (4*3*2)/(6*5*4)= 1/5
The player earnings would be: 3*$1.5 -$4= $0.5
wwwp= (4*3*2*2)/(6*5*4*3*2)= 2/15
The player earnings would be: 4*$1.5 -$4= $2
wwwwp= (4*3*2*2*1)/(6*5*4*3*2*1) = 1/15
The player earnings would be: 5*$1.5 -$4= $3.5
2. Find its expected value
The expected value would be:
chance of event * earning
You need to combine the 5 possible outcomes from the number 1 to get the total expected value.
Total expected value= (1/3 * - 2.5)+ (4/15*-1) + (1/5*0.5) + (2/15 *2) + ( 1/15 *3.5)=
(-12.5 -4 + 1.5 + 4 + 3.5) /15= -$7.5
This game basically a rip off.
Answer:
2.46 x 10⁵
Step-by-step explanation:
In 24 hours, the worm climbs 5 feet.
7-2= 5 (the worm falls 2 feet every night)
So,
5*8=40
It will take bruce's worm about 8 days to get to the top of the well.
I hope this helps!
~kaikers
Answer:
A?
Step-by-step explanation:
first off let's notice that the height is 11 meters and the volume of the cone is 103.62 cubic centimeters, so let's first convert the height to the corresponding unit for the volume, well 1 meters is 100 cm, so 11 m is 1100 cm.
![\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{cm^3}{103.62}\\ h=\stackrel{cm}{1100} \end{cases}\implies 103.62=\cfrac{\pi r^2 (1100)}{3} \\\\\\ 3(103.62)=1100\pi r^2\implies \cfrac{3(103.62)}{1100\pi }=r^2 \\\\\\ \sqrt{\cfrac{3(103.62)}{1100\pi }}=r\implies \stackrel{cm}{0.00510199305952} \approx r](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D%5Cstackrel%7Bcm%5E3%7D%7B103.62%7D%5C%5C%20h%3D%5Cstackrel%7Bcm%7D%7B1100%7D%20%5Cend%7Bcases%7D%5Cimplies%20103.62%3D%5Ccfrac%7B%5Cpi%20r%5E2%20%281100%29%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%203%28103.62%29%3D1100%5Cpi%20r%5E2%5Cimplies%20%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%3Dr%5E2%20%5C%5C%5C%5C%5C%5C%20%5Csqrt%7B%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Cstackrel%7Bcm%7D%7B0.00510199305952%7D%20%5Capprox%20r)