Assuming you have a cartridge with PS 81 on the back- all I can say is that it is a military cartridge, made in 1981. The PS code was used by 4 different countries, so can't really tell what cartridge that is.
p=1
7/8 P(all heads) = (1/2)*(1/2)*(1/2)=(1/8). P(at least one tails) = 1-P(all heads) = 7/8. Another (more time-consuming) way to get the same result: P(all tails) = (1/2)^3=1/8 P(exactly two tails) = 3*(1/2)^3=3/8 P(exactly one tails) = 3*(1/2)^3=3/8 P(at least one tail) = P(all tails) + P(exactly two tails) + P(exactly one tails) = 7/8
To find out how many times greater 8 is than 8p, you can divide 8 by 8p. This results in ( \frac{8}{8p} = \frac{1}{p} ). Therefore, 8 is ( \frac{1}{p} ) times greater than 8p, which means it depends on the value of p. If p is positive, 8 is greater than 8p when p is less than 1, and vice versa.
24 - 2001 Day 8 400 p-m--500 p-m- 8-1 is rated/received certificates of: Netherlands:12 USA:TV-14
13. Come on this one is easy. HAHAHA :P
The equation is as follows: p - 5/8 (1/4 + p). This equation expands to (1/4)p + p^2 - 5/32 - (5/8)p, which simplifies to p^2 - (3/8)p - 5/32. Now either solve for p using the quadratic formula or solve by factoring.
8-13 - 2010 814 P-M- 1-6 was released on: USA: 8 October 2010
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (7 plus or minus the square root of 145) divided by 16. p = 1.1900996611745185 p = -0.3150996611745185
C-O-P-S- - 1988 The Case of C-O-P-S- File 1 Part 1 1-8 was released on: USA: 21 November 1988
To find one third of the difference between 6 and ( p ), we first calculate the difference: ( 6 - p ). Then, one third of this difference is ( \frac{1}{3}(6 - p) ). When we add this to 8, the expression becomes ( 8 + \frac{1}{3}(6 - p) ). Thus, the final expression is ( 8 + \frac{6 - p}{3} ).
P-Dog's World - 2011 P-Dog's Mother's Day Special 1-8 was released on: USA: 24 May 2012
Consider the three events: A = rolling 5, 6, 8 or 9. B = rolling 7 C = rolling any other number. Let P be the probability of these events in one roll of a pair of dice. Then P(A) = P(5) + P(6) + P(8) + P(9) = 18/36 = 1/2 P(B) = P(7) = 6/36 = 1/6 and P(C) = 1 - [P(A) + P(B)] = 1/3 Now P(A before B) = P(A or C followed by A before B) = P(A) + P(C)*P(A before B) = 1/2 + 1/3*P(A before B) That is, P(A before B) = 1/2 + 1/3*P(A before B) or 2/3*P(A before B) = 1/2 so that P(A before B) = 1/2*3/2 = 3/4