README.md

BSI-reliability-calculator by Anastasiia Ponkratova, Julia Migiel

1. Calculation of reliability of each element in parallel in the system

    A  system is  composed of 5  identical independent elements in parallel. What should be the  reliability of each element to achieve a 
    reliability of 0.96 ?

    Answer:1-((1-0.96))^(1/5))=1-0,52=0,48

2. Comparison of the probability of A's success to that of B's success;

     A  has  one  share in  a  lottery in  which there is  one  prize and two blanks ;  B   has  three shares in  a  lottery in  which there are  three prizes and  6  blanks; compare the probability of A's success to that of B's success. 
    
     Answer:
     
     xCy= x!/(y!(x-y)!)
     
     Probability A to loss =6C3/9C3
     Probability A to win  = 1−9C3/6C3=1−6!/(3!*3!)×(3!*6!)/9!=1−120/504=384/504=16/21
     Probability B to loss = 2C1/3C1
     Probability B to win  = 1−(2C1/3C1)=1−2/3=1/3
     required ratio is
     16/21 : 1/3=16 : 7

3. Calculate probability that only one of woman or man will be selecte with specified probability

    A man and his wife appear in an interview for two vacancies in the same post. 
    The probability of husband's selection is 1/7 
    and the probability of wife's selection is 1/5 
    What is the probability that only one of them is selected ?

    Answer:
    P(A)=1/7
    P(B)=1/5
    
    So there is 6/7 probability for men to not to be selected and 4/5 for woman
    So required probability will be:
    P[(A and notB)or(B and notA)]=(1/7)*(4/5)+(1/5)*(6/7)=(4+6)/35=10/35

4. Calculation probability of finding 2 kids from group that contains x kids, y womans, z mens

  Four persons are  chosen at random from a group containing 3  men, 2  women and  4  children. 
  Calculate the chances that exactly two of them will be children.
  
   Answer:
   
   xCy= x!/(y!(x-y)!)
   
   All ways of groups:
   9C4=126
   Because we want 2 of choosen to be a children: 
   4C2=6
   and choose the other 2 people will give:
   5C2=10
   
   So the probability will be (10*6)/126 = 10/21

5. Finding probability of capacity availability in the cinema on any day

    A  cinema house gets electric power from a generator run  by diesel engine. On any day, 
    the probability that the generator is down (event A) is 0.025 and the probability 
    that the diesel engine is down (event B) is 0.04. What is the  probability that the 
    cinema house will have power on  any  given day? Assume that occurrence of 
    event A  and event B are independent of each other.
    
    Answer:
    
    At first we will calculate the probabilyty of losing power in cinema
    P(AUB)=P(A)+P(B)-P(A)*P(B)=0.025+0.04-0.025*0.04=0.064

    so the answer will be 1-0.064=0.936