### Scribe

Wednesday 8, 2006

Today is the last day we looked at probabilities. We only have 1 period today and Mr. K gave us some notes to write out in our Math dictionary so here it is.

P = Probabilities

A,B = Variables

* = multiplication

/ = division (fraction)

-Given any 2 events, A and B, the probability of A and B occuring is given by:

P(

NOTE: If A and B are MUTUALLY EXCLUSIVE (info about Mutually Exclusive event is in Mauxi's scribe) :

P(

-We calculate the probability of A or B as Follows:

P(

ex. Given a standard deck of cards (52 cards).

One card is drawn, Find:

i. P(J and H) = P(J) * P(H) = 4/52 * 13/52 = 1/52

4/52 = 4 jacks (Spades, heart, clover and diamond)

13/52 = ace

1/52 = jack of heart

4*13 = 52 52*52 = 2704

52/2704 / 52/52 = 1/52 (reduced)

ii. P(J or H) = P(J) + P(H) - P(J and H) = 4/52 + 13/52 - 1/52 = 16/52 (reduced) = 4/13

iii. P(J or 6) = P(J) + P(6) - P(J and 6) = is it possible?

The P(J) (probability of drawing a jack) and P(6) (probability of drawing a 6) are possible but drawing a JACK with 6 are not.

Therefore these are muatully exclusive.

Note: If the event is MUTUALLY EXCLUSIVE the probabality "AND" can't be used.

This is my scribe for today and hope that you got something from this.

The scribe for Friday will be Cait.

Today is the last day we looked at probabilities. We only have 1 period today and Mr. K gave us some notes to write out in our Math dictionary so here it is.

P = Probabilities

A,B = Variables

* = multiplication

/ = division (fraction)

**Calculating Probabilities**

Using "Using "

*And*" and "*OR*"-Given any 2 events, A and B, the probability of A and B occuring is given by:

P(

**and***A***) = P***B***(***A***)*** P*(***B***)*NOTE: If A and B are MUTUALLY EXCLUSIVE (info about Mutually Exclusive event is in Mauxi's scribe) :

P(

**and***A***) = 0***B*-We calculate the probability of A or B as Follows:

P(

**or***A***) = P***B***A***(***+ P***)**(B***- P (***)***and***A***)***B*ex. Given a standard deck of cards (52 cards).

**J**: Draw a jack**H**: Draw a heart**6**: Draw a 6One card is drawn, Find:

i. P(J and H) = P(J) * P(H) = 4/52 * 13/52 = 1/52

4/52 = 4 jacks (Spades, heart, clover and diamond)

13/52 = ace

**, 2h***H***, 3***H***, 4***H***... king***H***.***H*1/52 = jack of heart

4*13 = 52 52*52 = 2704

52/2704 / 52/52 = 1/52 (reduced)

ii. P(J or H) = P(J) + P(H) - P(J and H) = 4/52 + 13/52 - 1/52 = 16/52 (reduced) = 4/13

iii. P(J or 6) = P(J) + P(6) - P(J and 6) = is it possible?

The P(J) (probability of drawing a jack) and P(6) (probability of drawing a 6) are possible but drawing a JACK with 6 are not.

Therefore these are muatully exclusive.

Note: If the event is MUTUALLY EXCLUSIVE the probabality "AND" can't be used.

This is my scribe for today and hope that you got something from this.

The scribe for Friday will be Cait.

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