Question

A hat contains four balls. The balls are numbered 2, 4, 4, and 5. One ball is randomly selected

Answer 1

Answer:

C) If the sum is 6, a small cone will be ordered . the sum is 7 or 8 , a medium cone will be ordered. If the sum is 9, a large cone will be ordered

Step-by-step explanation:

If the sum is a 6: (2+4) or (4+2)

P(6) = (2 is first picked)(4 is picked next) + (4 is first picked)(2 is picked next)

$P(6)= (\frac{1}{4})(\frac{2}{3}) +(\frac{2}{4})(\frac{1}{3})$

$P(6) =\frac{2}{12} + \frac{1}{6} = \frac{1}{3}$

If the sum is a 7: (2+5) or (5+2)

P(7) = (2 is first picked)(5 is picked next) + (5 is first picked)(2 is picked next)

$P(7)= (\frac{1}{4})(\frac{1}{3}) +(\frac{1}{4})(\frac{1}{3})$

$P(7) =\frac{1}{12} + \frac{1}{12} = \frac{1}{6}$

If the sum is a 8: (4+4)

$P(8)= (\frac{2}{4})(\frac{1}{3})=\frac{1}{6}$

If the sum is a 9: (4+5) or (5+4)

P(9) = (4 is first picked)(5 is picked next) + (5 is first picked)(4 is picked next)

$P(9)= (\frac{2}{4})(\frac{1}{3}) +(\frac{1}{4})(\frac{2}{3})$

$P(9) =\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

$P(6) = \frac{1}{3}$

$P(7) = \frac{1}{6} ;P(8) = \frac{1}{6}$

$P(7) or P(8) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

$P(9) = \frac{1}{3}$

Therefore, if the sum is 6, a small cone will be ordered . the sum is 7 or 8 , a medium cone will be ordered. If the sum is 9, a large cone will be ordered