A factory produces biscuits using three machines: K, J, and L. The machines produce 45%, 30%, and 25% of the biscuits respectively. The probabilities of a biscuit being broken are 0.03, 0.02, and 0.05

• Machine K makes 45% of biscuits: $P(K) = 0.45$
• Machine J makes 30% of biscuits: $P(J) = 0.30$
• Machine L makes the rest: $P(L) = 1 - P(K) - P(J) = 1 - 0.45 - 0.30 = 1 - 0.75 = 0.25$

• For Machine J: $P(B|J) = 0.02$, so $P(B'|J) = 1 - 0.02 = 0.98$
• For Machine K: $P(B|K) = 0.03$, so $P(B'|K) = 1 - 0.03 = 0.97$
• For Machine L: $P(B|L) = 0.05$, so $P(B'|L) = 1 - 0.05 = 0.95$

                                      Start
                                        |
             -----------------------------------------------------
             |                          |                          |
             K (0.45)                   J (0.30)                   L (0.25)
             |                          |                          |
      -----------------          -----------------          -----------------
      |               |          |               |          |               |
      B (0.03)        B'(0.97)   B (0.02)        B'(0.98)   B (0.05)        B'(0.95)
      |               |          |               |          |               |
      P(K and B)      P(K and B') P(J and B)      P(J and B') P(L and B)      P(L and B')
      = 0.45*0.03     = 0.45*0.97 = 0.30*0.02     = 0.30*0.98 = 0.25*0.05     = 0.25*0.95
      = 0.0135        = 0.4365    = 0.0060        = 0.2940    = 0.0125        = 0.2375

The possible outcomes and their associated probabilities are:

• $P(\text{Machine K and Broken}) = 0.45 \times 0.03 = \boxed{0.0135}$
• $P(\text{Machine K and Not Broken}) = 0.45 \times 0.97 = \boxed{0.4365}$
• $P(\text{Machine J and Broken}) = 0.30 \times 0.02 = \boxed{0.0060}$
• $P(\text{Machine J and Not Broken}) = 0.30 \times 0.98 = \boxed{0.2940}$
• $P(\text{Machine L and Broken}) = 0.25 \times 0.05 = \boxed{0.0125}$
• $P(\text{Machine L and Not Broken}) = 0.25 \times 0.95 = \boxed{0.2375}$