Answer:

1.Let “S” denote a “spade” and “H” stand for “heart”.

Since all spades constitute for 13/52 of the deck then at any individual draw,

p(S)=1352=1/4.

The same applies for hearts.

Now, if one of the cards is drawn (e.g. let it be “S”), then probability that the next one is “H” follows under conditional probability:

p(H|S)=1352−1,

since the deck is one card less.

Now, we ask of the probability of such outcome when one is “H” and one is “S”, then:

p(H,S)=p(H|S)p(S)=135114=0,06372549.

The situation is completely symmetric, so p(H,S)=p(S,H).

To get the probability of such situation, when we draw two, but the order doesn’t matter, we have:

p(H\andS)=p(H,S)+p(S,H)=2×0.06372549=0,12745098.

2.There are 5 green and 7 red balls. If two balls are selected without replacement at random, then what is the probability that the first one is green and the second one is red?

This is nice and easy, because the two balls are being drawn in a specific order. One ball is drawn first, the other is drawn second.

The chance of the first ball being green is a simple 5/12. There are 5 green balls, but 12 total balls.

The chance of the second ball being red after the first one pulled was green is 7/11. There were 7 red balls, but 11 total balls to choose from.

Since both of these things must happen to satisfy our original statement (i.e. the statement is an “and” type statement), we simply multiply these chances together.

5/12*7/11=35/132.

Step-by-step explanation:

hipe its help

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