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ikadub [295]
3 years ago
7

You have also set up a card game in which a player picks a card from a standard deck of 52 cards. The player wins if these two e

vents occur together: E1, in which the card drawn is a black card, and E2, in which the card drawn is a numbered card, 2 through 10.What is the probability of getting a black card and a numbered card? Calculate the probabilities P(E1) and P(E2) as fractions.
Mathematics
1 answer:
Alenkinab [10]3 years ago
5 0

First, let's count:

there are 26 possible outcomes for E1 (black card)

there are 4x9 = 36 possible outcomes for E2, to pick a numbered card (any color)

there are 2x9 =18 possible outcomes for E1 (black) AND E2 (numbered, spade + clower)

the probability of E1 AND E2 is the ratio of the count of possible outcomes for E1 + E2 and the count of all possible outcomes (52 choices to pick a card from the deck):

P(E1 and E2) = 18/52 (34.6%)

And as asked:

P(E1) = 26/52 = 1/2 (50%)

P(E2) = 36/52 = 9/13 (69.2%)

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Zigmanuir [339]

Answer:

\frac{1}{2} (a + 2b)(a - b)

Step-by-step explanation:

Assuming you require the expression to be factored

Given

\frac{1}{2} a² + \frac{1}{2} ab - b² ← factor out \frac{1}{2} from each term

= \frac{1}{2} (a² + ab - 2b²) ← factor the quadratic

Consider the factors of the coefficient of the b² term(- 2) which sum to give the coefficient of the ab- term (+ 1)

The factors are + 2 and - 1, since

2 × - 1 = - 2 and 2 - 1 = + 1, thus

a² + ab - 2b² = (a + 2b)(a - b) and

\frac{1}{2} a² + \frac{1}{2} ab - b² = \frac{1}{2}(a + 2b)(a - b)

5 0
3 years ago
Help please !! 20 points
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