Bullseye --- I

The crux is to calculate the combined chance of missing bullseye

The probability of A,B C D respectively missing is 0.3, 0.4, 0.5 , 0.6

Multiplying them, combined chance of missing = 0.036

1- 0.036= 0.964 ( 96.4%) is the chance that at least one of them scores a bullseye.

probability that A, B, C, and D do hit is 7/10,6/10,5/10 and 4/10 respectively and the probability that they would miss is 3/10, 4/10, 5/10 and 6/10 respectively. probability that all would miss is (3x4x5x6)/(10x10x10x10) =360/10000 = 0.0360 = 3.6%. probablity that atleast one of them would hit is 1-(probability that none would hit) because 1 or 100% is total probability. answer - (1-0.036) or (100%-3.6%) = 0.964 or 96.4%

by inclusion exculsion , easily we get 1-(3/10)(4/10)(5/10)(6/10) = 0.964 (96.4%)

Probability that none of them hits the bullseye + Probability that at least one of them hitting bullseye = 1 Probabilities of not hitting bullseye for A,B,C,D are 0.7, 0.6, 0.5 and 0.4 respectively. Probability that none of them hits the bullseye = 0.7 0.6 0.5*0.4 = 0.036 Therefore, probability that at least one of them hitting bullseye (P) = 1-0.036 = 0.964 In percentage, P = 0.964 *100% = 96.4%

simple, 1 - (1 - 0.7) (1 - 0.6) (1 - 0.5) (1 - 0.4)

Getting the probability of atleast one is same as 1- probability of none. Probability of none = 3/10 * 4/10 * 5/10 * 6/10 = 0.036 Therefore probability of atleast one = 1-0.036=0.964 = 96.4 %

Using Principle of Inclusion Exclusion is a little bit messy here, I have done with De Morgan's law.

P(A U B U c U D)=1 - P(A' )P(B' )P(C' ) P(D' )

You do 1-0.3 0.4 0.5*0.6=0.964, in percnet it's 96.4

Probability of at LEAST 1 Hit = 1 - Probability of NO Hits at all = 1-(1-0.7) (1-0.6) (1-0.5)*(1-0..4) = 0.964 = 96.4%