Probability?

Box A, B and C are identical.





Box A contains 8 identical counters numbered 1,2,3,4,5,6,7 and 8.





Box B contains 6 identical counters numbered 2,3,5,7,8 and 9.





Box C contains 5 identical counters numbered 1,2,3,6 and 8.





A box is chosen at random and a counter is selected at random from the box.





Find the probability that





(a) Box B is chosen (ans: 1/3)





(b) Box C is chosen and the counter selected has an even number on it (ans: 1/5)





(c) Box A is chosen and the counter selected has a prime number on it, (ans: 1/6)





(d) counter selected has a prime number on it. (ans: 47/90)





Please show your workings clearly...thanks..

Probability?
q1


P=n(wanted box)/n(all boxes)


=1/3





q2


P=1/3 * n(C's even)/n(C)


=3/3*5


=1/5





q3


P=1/3 * n(A's prime)/n(A)


=4/3*8


=1/6





q4


P=1/3*[4/8 + 4/6 + 2/5]


=1/3*[1/2+2/3+2/5]


=[15+20+12]/2*3*5*3


=47/90
Reply:a) easy; 3 boxes total, Box B chosen. 1/3


b) ok for this u use the 1/3 from problem a) and multiply it to 3/5 becuase there are 3 even numbers out of 5


c) again use 1/3 from b) and a) and multiply to 4/8 or 1/2 becuase there are 4 primes out of 8 (2,3,5,7)


d) good luck with this i have no idea srry :)

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