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dreamer86
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#1
Report Thread starter 13 years ago
#1
Please may I have some help with this question,

In a game, each player throws four ordinary six-sided dice. The random variable x is the largest number showing on the dice.

i) Find the probability that X=1.

ii) Find the probability that X <2 and deduce that the probability that X=2 is 5/432.

iii) Find the probability that x = 3.

iv) Find the probability that x = 6 and explain without further calulation why 6 is the most likely value for x.

Any help will be appreciated, I can only do part i) at least i think i can, i get 1/1296, I did (1/6)4

Thankyou, my exam is thursday
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Jonny W
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#2
Report 13 years ago
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Your answer to (i) is correct.

(ii)
P(X <= 2)
= (1/3)^4
= 1/81

P(X = 2)
= P(X <= 2) - P(X = 1)
= 1/81 - 1/1296
= 5/432

(iii)
P(X <= 3)
= (1/2)^4
= 1/16

P(X = 3)
= P(X <= 3) - P(X <= 2)
= 1/16 - 1/81
= 65/1296

(iv)
P(X = 6)
= 1 - P(X <= 5)
= 1 - (5/6)^4
= 671/1296

Since P(X = 6) > 1/2, 6 is the most likely value for X.
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dreamer86
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#3
Report Thread starter 13 years ago
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(Original post by Jonny W)
Your answer to (i) is correct.

(ii)
P(X <= 2)
= (1/3)^4
= 1/81

P(X = 2)
= P(X <= 2) - P(X = 1)
= 1/81 - 1/1296
= 5/432

(iii)
P(X <= 3)
= (1/2)^4
= 1/16

P(X = 3)
= P(X <= 3) - P(X <= 2)
= 1/16 - 1/81
= 65/1296

(iv)
P(X = 6)
= 1 - P(X <= 5)
= 1 - (5/6)^4
= 671/1296

Since P(X = 6) > 1/2, 6 is the most likely value for X.

Thank you, how do you know what value it is to the power of 4? I have worked it out from your answers that for 1 its 1/6, 2 its 2/6 = 1/3, 3 its 3/6 = 1/2, 4 its 4/6 =2/3 and 5/6. but why? I know its to the power 4 beause there are 4 dice!
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Jonny W
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X is <= 2 if (and only if) all the dice are <= 2.

The probability of the first dice being <= 2 is 2/6 = 1/3.

The probability of all four dice being <= 2 is the fourth power of that probability: (1/3)^4.
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dreamer86
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#5
Report Thread starter 13 years ago
#5
aw thanks very much was having a little panicky moment
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