SOLUTION: A box contains 2 fifty paise coins,5 twenty five paise coins and a certain fixed number(N<=2) of ten and five paise
coins.If five coins are taken out of the box at random,then the
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Question 31598: A box contains 2 fifty paise coins,5 twenty five paise coins and a certain fixed number(N<=2) of ten and five paise
coins.If five coins are taken out of the box at random,then the probability that the total value of these 5 coins is
less than one rupee and fifty paise is which of the following:-
A.10*(N+2)/(N+7C5)
B.1-(10*(N+2)/(N+7C5))
C.5(N+3)/(N+5C2)
D.1-(5*(N+3)/(N+5C2))
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
A box contains 2 fifty paise coins,5 twenty five paise coins and a certain fixed number(N<=2) of ten and five paise
coins.If five coins are taken out of the box at random,then the probability that the total value of these 5 coins is
less than one rupee and fifty paise is which of the following:-
SINCE IT IS EASIER TO FIND THE NUMBER OF WAYS BY WHICH THE SUM OF 5 COINS CAN EQUAL OR EXCEED RS.1.50,LET US FIND IT OUT AND SUBTRACT THE RESULTANT PROBABILITY FROM ONE TO GET THE ANSWER
WE HAVE IN ALL 2+5+N=N+7 COINS
5 CAN BE TAKEN IN ...(N+7)C5 WAYS
CASE 1.....1*50+4*25=1.50....IN..(2C1)(5C4)=2*5=10
CASE 2.....2*50+3*25=1.75....IN..(2C2)(5C3)=1*10=10
CASE 3.....2*50+2*25+1*5 OR 10=1.55 OR 1.60...IN...(2C2)(5C2)(NC1)=1*10*N=10N
TOTAL NUMBER .....................=10+10+10N=20+10N
PROBABILITY OF GETTING 1.50 OR MORE =(20+10N)/{(N+7)C5}
PROBABILITY OF GETTING LESS THAN 1.50 =1- [(20+10N)/{(N+7)C5}]
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