Playing Game A and Playing Game B are independent events.
The probability Mr.lucky wins both games is 9/25 (0.36)
The probability that Mr. Lucky wins Game B is four times greater than him losing Game A
Find the probability that Mr. lucky wins only one of the two games.
i got 2.98 - but i think i am wrong
This question seems very hard for GCSE...however:
Let a and b respectively denote the probability that Mr.Lucky wins game a and game b. Can you use the information to find two simultaneous equations in a and b? I'll put the answers in a spoiler
Let a and b respectively denote the probability that Mr.Lucky wins game a and game b. Can you use the information to find two simultaneous equations in a and b? I'll put the answers in a spoiler
Spoiler
i understand what you did but instead i made a quadratic equation instead and yes i know its a hard question, my teacher gave us part of a grade 9
If my working out is correct your x value is incorrect. We have 2 ways of calculating the probability of winning A which we can then equate the solve for our unknown.
What you did seems reasonable. I like the tree diagram as it shows me what you are thinking. In your tree diagram, you are implicitly using the fact that the two games are independent, because you have 1-4x and 4x in BOTH of the 'game B' branches.
You then correctly obtained the quadratic 4x(1-x)=9/25, were x=P(lose game a)- but this quadratic has TWO roots. These are 0.1 and 0.9 (check this!) How can you tell which of these is the 'correct' one to take?
If my working out is correct your x value is incorrect. We have 2 ways of calculating the probability of winning A which we can then equate the solve for our unknown.
OP has made two separate threads, each containing this problem. It might be better to use just the other thread, which is here (I'll edit my post in a minute with the link):
What you did seems reasonable. I like the tree diagram as it shows me what you are thinking. In your tree diagram, you are implicitly using the fact that the two games are independent, because you have 1-4x and 4x in BOTH of the 'game B' branches.
You then correctly obtained the quadratic 4x(1-x)=9/25, were x=P(lose game a)- but this quadratic has TWO roots. These are 0.1 and 0.9 (check this!) How can you tell which of these is the 'correct' one to take?
i square rooted 0.16 and that would give +/- 0.4 i chose the positive value as it made more sense as you cant have a negative probability
If my working out is correct your x value is incorrect. We have 2 ways of calculating the probability of winning A which we can then equate the solve for our unknown.
In your working, you had +/-0.4 = x-0.5, so x=0.1 or 0.9. BOTH of these lie in the interval [0,1] so both can be probabilities. However, it turns out that only one of them is correct. Have a think about which one this might be.
In your working, you had +/-0.4 = x-0.5, so x=0.1 or 0.9. BOTH of these lie in the interval [0,1] so both can be probabilities. However, it turns out that only one of them is correct. Have a think about which one this might be.
ohh i see what you mean. i dont know what do you thnik?
I hope you've tried hard enough ahah, what I did is I let x=P(wins B) and then the probability of winning A is 0.36/x as well as 3x/4 (as the probability of losing a becomes x/4 so 1/(x/4)=3x/4 )