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Kisachek [45]
4 years ago
14

Suppose you have a bag containing 15 red beads, 12 white beads and 8 black beads. You are going to draw one bead out of the bag.

What is the probability that you draw a black bead or a white bead. Give your answer as a reduced fraction.
Mathematics
1 answer:
iren [92.7K]4 years ago
8 0

Answer:

probability that you draw a black bead or a white bead that is P(E1 U E2)

= 20/35 = 0.571

Step-by-step explanation:

Given a bag containing 15 red beads, 12 white beads and 8 black beads

n(S) = 15 + 12 +8 = 35

Let 'E1 be the event of selecting black beads and E2 be the event of selecting white beads

n(E1) = 8 and n(E2) = 12

Probability of draw a black bead P( E1 ) = \frac{n(E_{1} )}{n(S)}

                                                      P(E_{1}) = \frac{8}{35}

Probability of draw a white bead P( E2 ) = \frac{n(E_{2} )}{n(S)}

                                                    P(E_{2}) = \frac{12}{35}

probability that you draw a black bead or a white bead that is P(E1 U E2)

and E1 n E2 = ∅ (disjoint events)

<u>Axiom of union </u>

<u></u>P(E_{1}UE_{2} ) = P(E_{1})+P(E_{2}) - P(E_{1}nE_{2})\\<u></u>

E1 n E2 = ∅ ⇒ P(E1 n E2) = p(∅) = 0

P(E_{1}UE_{2} ) = P(E_{1})+P(E_{2}) - 0

P(E_{1}UE_{2} ) = \frac{8}{35} +\frac{12}{35}

P(E_{1}UE_{2} ) = \frac{20}{35}

P(E_{1}UE_{2} ) = 0.571

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