a man has 3 unique coins such that the probability of obtaining a head when the coin is flipped is 1 10 , 2 10 and 3 10 , respectively. If he flips each of the 3 coins once, the probability that at least 1 heads appears is p . What is the value of 1000 p ?
Good grief
You first posted this on Wednesday
and I replied you this way:
http://www.jiskha.com/display.cgi?id=1360139828
Then you reposted exactly the same way,
bobpursley gave you a reply assuming a typing error
http://www.jiskha.com/display.cgi?id=1360487406
Now you are posting it again, and behold, it again makes no sense
Did you mean 1/10 , 1/10 etc ????
If so just make the necessary changes in bob's solution.
Well............u cheat quite well.Should I report this to the authorities? :p
you maths freak talking to your father liske thissssssss
To find the probability that at least one heads appears when flipping each of the 3 coins once, we can use the concept of complement probability.
The probability of obtaining a tails when flipping a coin once is given by 1 minus the probability of obtaining a heads. Let's calculate the probability of obtaining a tails for each of the 3 coins:
First coin: Probability of tails = 1 - (1/10) = 9/10
Second coin: Probability of tails = 1 - (2/10) = 8/10
Third coin: Probability of tails = 1 - (3/10) = 7/10
Since each coin is flipped independently, the probability of obtaining a tails on all 3 coins is calculated by multiplying the individual probabilities:
Probability of tails on all 3 coins = (9/10) * (8/10) * (7/10) = 504/1000 = 0.504
Now, we can find the probability that at least one heads appears by subtracting the probability of obtaining tails on all 3 coins from 1:
Probability of at least one heads = 1 - 0.504 = 0.496
To find the value of 1000p, we multiply the probability by 1000:
1000p = 1000 * 0.496 = 496
Therefore, the value of 1000p is 496.