a game is played by throwing a fair dice until either a 6 is obtained or 4 throws have been made.?

Question

The dice is thrown and then an unbiased coin is thrown the number of times indicated by the score on the dice. Let H denote the number of heads obtained.Show that P(H) = 13/48

Answer 1

Are you sure? Unless I'm misreading this, you're suggesting that the number of heads is not a whole number, regardless of how many tosses or throws you've made...

A start, then is consider the possiblity that no 6s are thrown on a first roll, this occurs with probability (5/6), the average throw is 3, and so the average number of heads is 1.5...

Answer 2

As posted, the question doesn't make sense. "The probability of the number of heads equals 13/48"? Something's missing.

Let's figure out the odds anyway. I have time to burn, and the problem amuses me.

First, let's figure out how many coin flips you get.
Note that any die roll of 6 gives you 6 flips. Any non-6 die roll will give you, on average, 3 flips (the average of 1, 2, 3, 4, 5).

The odds of the first roll being a 6 are 1/6.
This gives you a 1/6 chance at getting 6 flips.
Expected number of flips from this scenario: 1, or 1296/1296

The odds of the second roll being a 6 are (5/6) * (1/6).
This gives you a 5/36 chance at getting 9 flips (3 expected flips from the first die, 6 from the second).
Expected number of flips from this scenario: 45/36, or 1620/1296

The odds of the third roll being a 6 are (5/6)*(5/6)*(1/6).
This gives you a 25/216 chance at getting 12 flips.
Expected number of flips from this scenario: 300/216, or 1800/1296

The odds of the fourth roll being a 6 are (5/6)*(5/6)*(5/6)*(1/6).
This gives you a 125/1296 chance at getting 15 flips.
Expected number of flips from this scenario: 1875/1296

The odds of four non-six rolls are (5/6)*(5/6)*(5/6)*(5/6).
This gives you a 625/1296 chance at getting 12 flips.
Expected number of flips from this scenario: 7500/1296

Total expected flips: 14091/1296, or roughly 10.87

The expected number of heads is half that, or 5.44.

I have no idea what the 13/48 in the question is supposed to be.

Answer 3

Do you mean E(H) = 13/48. The expected number of heads? This would certainly be more than 1.

Answer 4

A proper fraction for sure
ie A number between 1 and 0

Answer 5

First you enumerate all obtainable outcomes and calculate the danger, Pi, the place i is the style of sixes, i = one million, 2, 3, 4, 0. it is the danger distribution. You calculate the propose of that distribution because of the fact the sum of the made of Pi circumstances i. then you calculate the variance interior an identical way different than you sum Pi * i^2. then you take the sq. root of that to get the common deviation. i do no longer understand no remember if or no longer they choose you to apply the biased or impartial form of the common deviation formula. The ananlysis of the style of rolls is comparable different than #sixes = 0 is replaced by making use of #rolls = 4. There are a optimal style of four throws. The 6 would be rolled on roll one million, 2, 3, or 4, or on no roll (0). enable the danger of the style of sixes be denoted P1, P2, P3, P4, P0. Then P1 = one million/6. P2 is the danger of rolling one million-5 on roll1 and then rolling a 6. it is (5/6) * (one million/6). further, P3 = (5/6)^2*(one million/6), and P4 = (5/6)^3*(one million/6). ultimately, P0 = (5/6)^4. So the predicted style of rolls is the sum of the products of the style of rolls circumstances the danger of that style of rolls: E{#sixes} = one million*P1 + 2*P2 + 3*P3 + 4*P4 + 0*P0 = one million/6 + (5/6)*(one million/6) + (5/6)^2*(one million/6) + (5/6)^3*(one million/6) = (one million/6)* (one million + 5/6 + (5/6)^2 + (5/6)^3) = one million.1775 Var{#sixes} = one million*P1 + 4*P2 + 9*P3 + sixteen*P4 + 0*P0 - E{#sixes}^2 = one million.92066 StdDev{#sizes} = sqrt(Var) = one million.38588 further, the prognosis of the style of throws provides: E{#throws} = one million*P1 + 2*P2 + 3*P3 + 4*P4 + 4*P0 = one million/6 + (5/6)*(one million/6) + (5/6)^2*(one million/6) + (5/6)^3*(one million/6) + (5/6)^4 = 3.1065 Var{#throws} = one million*P1 + 4*P2 + 9*P3 + sixteen*P4 + 4*P0 - E{#throws}^2 = one million.3729 StdDev{#throws} = sqrt(Var) = one million.17172 for this reason the predicted style of sixes given above (0.5177) is inaccurate. you're able to have the skill to persist with an identical prognosis to get carry of the predicted style of things after one sport.

Answer 6

Why, all things are possible in probability. The rules say so. The likelihood of it being so is another matter.