Heads I win, Tails you lose

According to given probability of obtaining a head, the probability of obtaining a tail when the coin is flipped is $\frac {9}{10}, \frac {8}{10}$ and $\frac {7}{10}$ , respectively. So if he flips each of the 3 coins once, the probability that all appears as tail is $\frac {9 \times 8 \times 7}{10 \times 10 \times 10}$ which is equal to $\frac {504}{1000}$ . So the probability that at least 1 heads appears is $\frac {1000-504}{1000}$ which is equal to $\frac {496}{1000}$ . So $1000p$ is equal to $496$ .

[latex edits]

We use the complement method.

The probability that at least 1 head appears is equivalent to the probability that we do not obtain all 3 tails (or whatever the opposite of a head is).

To calculate $$P(\text{3 tails})$$ we simply take $$(1-\frac1 {10}) \cdot (1-\frac2 {10}) \cdot (1-\frac3 {10})=\frac{9\cdot8\cdot7}{1000}=\frac{504}{1000}$$ assuming that the 3 unique coins are independent of each other.

As our goal is to find the probability that we do not obtain all 3 tails, we subtract this from 1, yielding $$p=1-\frac{504}{1000}\Rightarrow 1000p=496$$

Since all the events are independent, we can apply the product rule to get the answer.

There will be no head in the result, only if we get all tails.

According to the product rule, the probability of getting all tails are ->

(1 - 1/10) * (1 - 2/10) * (1 - 3/10) = 504/1000

Therefore the required result is (1 - 504/1000) = 496/1000

We count the probability of 0 head appear. then we use P(A)=1-P(A'). The probility of 0 head appear is P(A')=(1-1/10)(1-2/10)(1-3/10)=504/1000. So we get P(A)=1-504/1000=496/1000=p. 1000p=496.

we have to find the probability of getting atleast 1 head so we can get 2 heads and 3 heads also but atleast 1 heads. the probailties of each case are given as 1/10,2/10,3/10. p(a)=probability of getting head in first case=1/10 p(b)=probability of getting head in second case=2/10 p(c)=probability of getting head in third case=3/10 and p'(a),p'(b),p'(c) are probability of not getting head in respective cases case 1:all three heads =p(a).p(b).p(c) =6/1000 case 2:two head =p(a).p(b).p'(c)+p'(a).p(b).p(c)+p(a).p'(b).p(c) =(14+54+24)/1000 =92/1000 case 3 :1head =p(a).p'(b).p'(c)+p'(a).p'(b).p(c)+p'(a).p'(b).p(c) =(56+126+216)/1000 =398/1000 adding all up total probability p =496/1000 therefore 1000p=496

There are three cases where heads appears at least once. When there are three, two, and one head(s) flipped.

Three heads: (1/10)(2/10)(3/10) Two heads: (1/10)(2/10)(7/10)+(1/10)(8/10)(3/10)+(9/10)(2/10)(3/10) One head: (1/10)(8/10)(7/10)+(9/10)(2/10)(7/10)+(9/10)(8/10)(3/10)

Adding all of these together gets you 496/1000 so 1000p=496.

The question asked "at least 1 heads appear", we just need to count the probability that no head appears, that is $\frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} = \frac{504}{1000}$ . So the value of $p$ is $1 - \frac{504}{1000} = \frac{496}{1000}$ . The answer is $1000p=496$ .

occurence of tail/head on one coin is independent of occurence of tail/head on other coin. P(atleast one head)=1-P(head on niether of coins)=1-P(tails on all coins) \Rightarrow P(atleast one head)=1-P((tail on coin 1)\cap(tail on coin 2)\cap(tail on coin 3)) \RightarrowP(atleast one head)=1-P(tail on coin 1)\timesP(tail on coin 2)\timesP(tail on coin 3) \RightarrowP(atleast one head)=1-(9/10)\times(8/10)\times(7/10)=1-504/1000 \RightarrowP(atleast one head)=p=496/1000 \Rightarrow1000p=496

First we find the probability that no heads appears and minus that from 1. That is the equivalent of getting all tails which is just $\frac{9}{10}*\frac{8}{10}*\frac{7}{10} = \frac{504}{1000} \Rightarrow 1 - \frac{504}{1000} = \frac{496}{1000} \Rightarrow 1000*\frac{496}{1000} = 496$

Since we are asked of the probability of at least 1 heads appearing, then it is easier if we just get first the probability that no heads will appear and then subtract it from 1. probability that no heads will appear is: (1 - 1/10)(1 - 2/10)(1 - 3/10) = 504/1000 probability that at least 1 heads will appear: 1 - 504/1000 = 496/1000 therefore, 1000p = 496

In first coin probability of not getting head = $\frac {9}{10}$ In second coin probability of not getting head = $\frac {8}{10}$ In third coin probability of not getting head = $\frac {7}{10}$ So when the three coins are flipped together, The probability of not getting head = $\frac {9*8*7}{10*10*10}$ = $\frac {504}{1000}$ So the probability of getting atleast one head=1- $\frac {504}{1000}$ = $\frac {496}{1000}$ So the probability of getting atleast one head (p)=0.496 So 1000p=0.496*1000=496

p(getting all tails)=(1-1/10)(1-2/10)(1-3/10) =504/1000 p(getting at least one head)=1-504/1000=496/1000 so 1000p=496

Let the coins be named A, B, C. The probability of obtaining a head from coins A, B, and C are respectively 1/10, 2/10, and 3/10.

Let P be the event of at least 1 appearance of a head. Consider Pc (the complement of P, i.e. the event of no heads appearing).

For Pc to occur... The outcome of coin A must be tails. This probability is 1-1/10= 9/10. The outcome of coin B must be tails. This probability is 1-2/10= 8/10. The outcome of coin C must be tails. This probability is 1-3/10= 7/10. Multiplying these, we get that the probability of Pc is 9 8 7/10^3= 504/1000.

Then, p= probability of P= 1-504/1000= 496/1000

So, 1000p= 496

The question asks for the probability of getting at least 1 head. So we have to just subtract the probability of getting 3 Tails from 1.

That is, P(Getting at least 1 head) = 1 - P(Getting 3 Tails)
This holds because every case other than 3 Tails will contain 2 Tails(i.e, 1 Head) ; 1 Tails (i.e, 2 Heads) and 0 Tails (i.e, 3 Heads) only. And that is what we want.
Now, Probability of getting Tails in the three coins are:
1st coin = $\frac {10 - 1}{10} = \frac {9}{10}$
2nd coin = $\frac {10 - 2}{10} = \frac {8}{10}$
3rd coin = $\frac {10 - 3}{10} = \frac {7}{10}$

Hence, Probability of getting 3 Tails = $\frac {9}{10} \times \frac {8}{10} \times \frac {7}{10} = \frac {504}{1000}$ .
Now, P(Getting at least 1 head) = 1 - P(Getting 3 Tails)
$\Rightarrow$ P(Getting at least 1 head) = 1 - $\frac {504}{1000} = \frac {1000 - 504}{1000} = \frac {496}{1000}$
Now, p = $\frac {496}{1000} \times 1000$
$\Rightarrow$ p = 496

Therefore, the answer is 496

The sum of the probabilities that at least 1 heads appear and that no heads appear is 1 which means that one of them will surely happen. To get the probability that at least 1 heads appear, it is better to simply compute for the probability that no heads will appear and simply subtract it to 1. Doing so, $\frac{10 - 1}{10} \times \frac{10 - 2}{10} \times \frac{10 - 3}{10} = \frac{504}{1000}$ .

Hence, the probability $p$ is $\frac {1000 - 504}{1000} = \frac{496}{1000}$ and $1000p = 496$ .

Probability of obtaining a head (H) when first coin is flipped: 1/10

Probability of obtaining a tail (T) when first coin is flipped: 1-1/10=9/10

Probability of obtaining H when second coin is flipped: 2/10

Probability of obtaining T when second coin is flipped: 1-2/10=8/10

Probability of obtaining H when third coin is flipped: 3/10

Probability of obtaining T when third coin is flipped: 1-3/10=7/10

Probability of obtaining at least 1 H if each of the 3 coins are flipped once: p

p = 1 - Probability of obtaining 0 H

= 1 - Probability of obtaining 3 T

= 1 - (9/10 * 8/10 * 7/10)

= 1 - (504/1000)

p = 496/1000

1000*p = 496

As we know, the probability of obtaining head when the coin is flipped is \frac {1}{10},\frac {2}{10},\frac {3}{10} respectively, so : The probability of not obtaining heads is \frac {9}{10},\frac {8}{10},\frac {7}{10} respectively

Let P(A) be the probability that at least 1 heads appears ; so P^C(A) be the probability that heads never appear

P(A) = 1 - P^C(A)

p = 1 - \frac {9}{10}.\frac {8}{10}.\frac {7}{10}=1-\frac {504}{1000}

p = \frac {496}{1000}

So : 1000p= 496

Possibility that at least 1 heads appear: p=1- ( (9/10) * (8/10) * (7/10) ) =496/1000 So: 1000p = 496

Set the 3 coins as A,B and C.If A has a probability of obtaining a head that is $\frac {1}{10}$ ,then the probability of tails is $\frac {9}{10}$ .For B,the probability of tails is $\frac {8}{10}$ .For C,the probability of tails is $\frac {7}{10}$ .Then we multiply $\frac {9}{10}$ by $\frac {8}{10}$ by $\frac {7}{10}$ which equals $\frac {504}{1000}$ .Remember that this is the probability that they are all tails,so the probability that at least 1 head appears is 1- $\frac {504}{1000}$ = $\frac {496}{1000}$ .So P is $\frac {496}{1000}$ and 1000P is 496.

1 head 1/10x8/10x7/10=56/1000 9/10x2/10x7/10=126/1000 9/10x8/10x3/10=216/1000

2heads 1/10x2/10x7/10=14/1000 1/10x8/10x3/10=24/1000 2/10x3/10x9/10=54/1000

3heads 1/10x2/10x3/10=6/1000

p= (56+126+216+14+24+54+6)/1000 =496/1000 1000p=1000x496/1000 = 496

It is easier to approach this problem by counting the complement. Let $X$ be the number of heads that appear after all the coins are flipped. We have $p = P(X \geq 1) = 1 - P(X = 0)$ .

To get $0$ heads, all of the coins must show a Tail. Thus $P(X=0) = \left(1 - \frac{1}{10} \right)\left(1 - \frac{2}{10} \right) \left(1 - \frac{3}{10} \right) = \frac{504}{1000}$ .

Therefore $p = P(X\geq 1) = 1 - \frac{504}{1000} = \frac{496}{1000}$ . Hence $1000p = 496$ .