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Two S1 questions!! Big help needed!!
10 At a fair, a roll-a-penny stall can be played with 1p or 2p coins. If the coin lands inside a square (without touching the edges) the player receives the coin plus 2 other coins of the same value, otherwise the coin is lost. The probability of winning the prize with a 1p coin is 19/40 and the probability for a 2p coin is 11/40.
a) Find the expected winnings for each coin.
b) Would you play this game and why?
c) The stall was eventually closed down by the management. Give a possible reason for this.
12 A box A contains 9 red balls and one white ball. A box B contains 8 red balls and one white ball. A ball is to be taken at random from box A and put into box B, and then a ball is to be taken at random from box B. Find the probability that this ball from box B will be white.
Of the 2 balls drawn, one from A and one from B, let X denote the number that are white. Find the probability distribution of X.
Find the mean of X.
Find also, to 2 decimal places, the variance of X.
Muchos Gracias in advance, Amigos!
a) Find the expected winnings for each coin.
b) Would you play this game and why?
c) The stall was eventually closed down by the management. Give a possible reason for this.
12 A box A contains 9 red balls and one white ball. A box B contains 8 red balls and one white ball. A ball is to be taken at random from box A and put into box B, and then a ball is to be taken at random from box B. Find the probability that this ball from box B will be white.
Of the 2 balls drawn, one from A and one from B, let X denote the number that are white. Find the probability distribution of X.
Find the mean of X.
Find also, to 2 decimal places, the variance of X.
Muchos Gracias in advance, Amigos!
(10)
(a)
1p: E(winnings) = (19/40)*3 = 57/40
2p: E(winnings) = (11/40)*6 = 33/20
(b)
Yes, using 1p coins - because the expected winnings are greater than the stake.
(c)
Lots of people played using 1p coins.
--
(12)
Draw a tree diagram.
P(ball from box B is white) = (9/10)(1/10) + (1/10)(2/10) = 11/100
P(X = 0) = (9/10)(9/10) = 81/100
P(X = 2) = (1/10)(2/10) = 2/100
P(X = 1) = 1 - 81/100 - 2/100 = 17/100
E(X) = 17/100 + 2(2/100) = 21/100
E(X^2) = 17/100 + 4(2/100) = 1/4
Var(X) = E(X^2) - E(X)^2 = 1/4 - (21/100)^2 = 0.21
(a)
1p: E(winnings) = (19/40)*3 = 57/40
2p: E(winnings) = (11/40)*6 = 33/20
(b)
Yes, using 1p coins - because the expected winnings are greater than the stake.
(c)
Lots of people played using 1p coins.
--
(12)
Draw a tree diagram.
P(ball from box B is white) = (9/10)(1/10) + (1/10)(2/10) = 11/100
P(X = 0) = (9/10)(9/10) = 81/100
P(X = 2) = (1/10)(2/10) = 2/100
P(X = 1) = 1 - 81/100 - 2/100 = 17/100
E(X) = 17/100 + 2(2/100) = 21/100
E(X^2) = 17/100 + 4(2/100) = 1/4
Var(X) = E(X^2) - E(X)^2 = 1/4 - (21/100)^2 = 0.21
Nufc_2005
10 At a fair, a roll-a-penny stall can be played with 1p or 2p coins. If the coin lands inside a square (without touching the edges) the player receives the coin plus 2 other coins of the same value, otherwise the coin is lost. The probability of winning the prize with a 1p coin is 19/40 and the probability for a 2p coin is 11/40.
a) Find the expected winnings for each coin.
b) Would you play this game and why?
c) The stall was eventually closed down by the management. Give a possible reason for this.
a) Find the expected winnings for each coin.
b) Would you play this game and why?
c) The stall was eventually closed down by the management. Give a possible reason for this.
1p:
P(W = 2) = 21/40
P(W = -1) = 21/40
E(W) = ∑xP(W = x) = (2)(19/40) + (-1)(21/40) = (38-21)/40 = 15/40 = 0.375p
2p:
P(W = 4) = 11/40
P(W = -2) = 29/40
E(W) = ∑xP(W = x) = (4)(11/40) + (-2)(29/40) = (44-58)/40 = -14/40 = -0.21p
b) Yes, because the odds are in your favor in the 1p version.
c) It was losing money because people were playing the 1p version and winning loads.
(I am the only one who hates the waffle questions like b) and c) on a maths paper?)
meathead
1p:
P(W = 2) = 21/40
P(W = -1) = 21/40
E(W) = ∑xP(W = x) = (2)(19/40) + (-1)(21/40) = (38-21)/40 = 15/40 = 0.375p
2p:
P(W = 4) = 11/40
P(W = -2) = 29/40
E(W) = ∑xP(W = x) = (4)(11/40) + (-2)(29/40) = (44-58)/40 = -14/40 = -0.21p
b) Yes, because the odds are in your favor in the 1p version.
c) It was losing money because people were playing the 1p version and winning loads.
(I am the only one who hates the waffle questions like b) and c) on a maths paper?)
P(W = 2) = 21/40
P(W = -1) = 21/40
E(W) = ∑xP(W = x) = (2)(19/40) + (-1)(21/40) = (38-21)/40 = 15/40 = 0.375p
2p:
P(W = 4) = 11/40
P(W = -2) = 29/40
E(W) = ∑xP(W = x) = (4)(11/40) + (-2)(29/40) = (44-58)/40 = -14/40 = -0.21p
b) Yes, because the odds are in your favor in the 1p version.
c) It was losing money because people were playing the 1p version and winning loads.
(I am the only one who hates the waffle questions like b) and c) on a maths paper?)
I agree! Maths is about lazy and using weird notation and formulae- Use of any English Language should be forbidden!
Oh dear...I'm got my S1 maths exam next thursday and I found these questions really hard! I want to get a good A as well! Better do some more revision 
MissSurfer
Oh dear...I'm got my S1 maths exam next thursday and I found these questions really hard! I want to get a good A as well! Better do some more revision
Back to the drawing board, eh MissSurfer? I got a C at GCSE and a C in my AS-level, but I'm aiming for an A because I'm that damn good.
Nufc_2005
Back to the drawing board, eh MissSurfer? I got a C at GCSE and a C in my AS-level, but I'm aiming for an A because I'm that damn good.
Haha Good Luck...you can do it!
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