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Fittoniya [83]
3 years ago
8

For each of these compound propositions, use the conditional-disjunction equivalence (Example 3) to find an equivalent compound

proposition that does not involve conditionals.
(a) ~p→ ~q
(b) (p v q) → ~p
(c) (p→ ~q) → (~p →q)
Mathematics
1 answer:
VashaNatasha [74]3 years ago
5 0

Answer:

(a) p v ~q

(b) ~(p v q) v ~p

(c) ~(~p v ~q) v (p v q)

Step-by-step explanation:

The conditional-disjunction equivalence is:

P→Q ⇔ ~P v Q

To find an equivalent compound proposition without the conditionals (without the "→") you have to apply the previous equivalence and simplify if possible.

a) ~p→~q

In this case, P= ~p and Q= ~q

Applying the equivalence:

~(~p) v ~q

p v ~q

b) (p v q) → ~p

In this case P = (p v q) and Q= (~p)

Applying the equivalence:

~(p v q) v ~p

c) (p→~q) → (~p→q)

In this case, you have to apply the conditional-disjunction equivalence for every conditional in the compound proposition.

First, let P= (p→~q) and Q= (~p→q)

~ (p→~q) v (~p→q)      (1)

Now, you have to find an equivalent compound proposition for both (p→~q) and (~p→q)

For (p→~q):

Let P= p and Q=~q

~p v ~q    

For (~p→q)

Let P= ~p and Q= q

~(~p) v q

p v q    

Then the expression (1) is:

~(~p v ~q) v (p v q)

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Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

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