Statistics question - probability (it is long but help on even just 1 question would be grateful)?

A fairground game involves trying to hit a moving target with a gun. A round consists of up to 3 shots. ten points are scored for a hit target, but the round is over if the player misses. Linda has a constant probability of 0.6 of hitting the target and shots are independant of one another.





a) Find the probability that Linda scores 30 points in a round





The random variable X is the number of points Linda scores in a round





b) Find the probability distribution of X





c) Find the mean and the stnadard deviation of X





A game consists of 2 rounds





d) Find the probability that Linda scores more points in round 2 than in round 1

Statistics question - probability (it is long but help on even just 1 question would be grateful)?
a) 0.6*0.6*0.6=0.216





b)


P(0)=0.4


P(10)=0.6*0.4=0.24


P(20)=0.6*0.6*0.4=0.144


P(30)=0.6*0.6*0.6=0.216





c)


Mean=0*0.4+10*0.24+ 20*0.144+30*0.216= 11.76





Expected value of X^2=0^2*0.4+10^2*0.24+ 20^2*0.144+30^2*0.216= 276





Variance=Expected value of X^2-Mean^2=176-11.76^2=137.7





or you can get the variance a different way that doesn't involve the expected value of X^2:


Variance=(0-11.76)^2*0.4+ (10-11.76)^2*0.24+ (20-11.76)^2*0.144+ (30-11.76)^2*0.216= 137.7





Standard deviation= Variance^0.5= 137.7^0.5=11.73





d)


Add up the probability of any of the following scenarios:


Score 0 in round 1, score 10 or above in round 2: Probability 0.4*(0.24+0.144+0.216)=0.24


Score 10 in round 1, score 20 or above in round 2: Probability 0.24*(0.144+0.216)=0.0864


Score 20 in round 1, score 30 in round 2: Probability 0.144*0.216=0.031104





0.24+0.0864+0.031104=0.357504. And that's the final answer.





Note that alternatively, you could've gotten the first part as:


Score 0 in round 1, score 10 or above in round 2: Probability 0.4*(1-0.4)=0.24





You can use similar alternate methods for the other parts, although it doesn't save much time.











Note that if you want, there's another method you could use for all of part d. Get the probability of getting the same number of points in round 1 and round 2:


Add up the probability of any of the following scenarios:


Score 0 in round 1 and 0 in round 2: Probability 0.4*0.4=0.16


Score 10 in round 1 and 10 in round 2: Probability 0.24*0.24=0.0576


Score 20 in round 1 and 20 in round 2: Probability 0.144*0.144=0.020736


Score 30 in round 1 and 30 in round 2: Probability 0.216*0.216=0.046656





0.16+0.0576+0.020736 +0.046656=0.284992.





Then the probability of getting a different number of points in round 1 than round 2 is 1 minus that=1-0.284992=0.715008.





Since they have the same distribution, if you know that you got a different number of points in round 1 than round 2, then half the time round 1 will have more points than round 2, and the other half of the time round 2 will have more points than round 1. So the final answer is 0.715008/2=0.357504.
Reply:A) .216 or 21.6 percent

quince