7B combining probabilities
Use the "at least once" rule to find the probabilities of the following event. Getting at least one head when tossing nine fair coins
1 - (1/2) ^ 9 = 511/512
Getting rain at least once in 4 days if the probability of rain on each single day is 0.5
1 - (1/2) ^4 = 0.938
Spinning two winners in a row with a wheel of fortune on which the winner is one of 29 equally likely outcomes.
1/29 x 1/29 = 1/841
Suppose you roll a die 6 times. What is the probability of getting at least one odd number?
3/6 = 1/2 (1/2) ^6 = 1/64 1- 1/64 = 63/64
Drawing at least one ten when you draw a card from a standard deck 4 times (replacing the card each time you draw)
4 tens in 52 cards 48 cards aren't 10s P(not A) = probability of drawing a card that is not a 10 = 12/13 1 - (12/13)^4 = 0.2740
Drawing three jacks in a row from a standard deck of cards when the drawn card is not returned to the deck each time
4/52 x 3/51 x 2/50 = 1/5225
suppose you roll a single die, what is the probability of rolling either a 2 or a 3
P (2 or 3) = P (2) + P(3) = 1/6 + 1/6 = 2/6 =1/3
a 3 person jury must be selected at random from a pool that has 6 men and 6 women. what is the probability of selecting an all male jury
P (all male)= P (1st is male, 2nd is male, 3rd is male) = 6/12 x 5/11 x 4/10 = 120/1320 = 0.091
In which of the following situations are the events non-overlapping? Select all that apply.We roll a die, hoping for a 2 or a 5.
We roll a die, hoping for a 2 or a 5.
Randomly drawing and immediately eating two red pieces of candy in a row from a bag that contains 10 red pieces of candy out of 43 pieces of candy total.
dependent 10/43 x 9/42 = 0.050
Drawing three red cards in a row from a standard deck of cards when the drawn card is not returned to the deck each time
dependent, 2/17 ( 0.1176)
What is the probability of getting a parking ticket on campus on at least one out of 6 days without a parking pass if the chance of not getting a ticket on a particular day when a parking pass isn't present is 0.15?
(0.15)^6 = 0.000 1 - 0.000 = 100%
use the at least once rule to find the probability of at least one head when you toss three coins
P (at least one head) = 1 - P (no head) = 1 - (1/2)(1/2)(1/2) = 7/8
you purchase 10 lottery tickets with the probability of winning 1 in 10. what is the probability that you will have at least 1 winning ticket among the ten
P (at least one wins) = 1 - P (none win) = 1 - (9/10)^10 =0.651
Purchasing at least one winning lottery ticket out of 10 tickets when the probability of winning is 0.04 on a single ticket
P (no A in one trial) = 1 - 0.04 = 0.96 1 - 0.96^10 = 0.3352
what is the probability of rolling three 4's in a row with a single die?
P(4 on first roll , 4 on second roll, 4 on third roll) = P(4 on 1st roll) x P(4 on 2nd roll)x P(4 on 3rd roll) = 1/6 x 1/6 x 1/6 = 1/216
Use the "at least once" rule to find the probability of getting at least one 6 in four rolls of a single fair die.
P(6) = 1/6 P(not 6) = 1 - P(6) = 1 - 1/6 = 5/6 1 - (5/6) ^ 4 = 0.518
If A and B are non-overlapping events, then
P(A or B) = P(A) + P(B)
If A and B are overlapping events, then
P(A or B) = P(A) + P(B) - P(A and B)
what is the probability of drawing 2 aces from a deck of cards
P(A1) x P(A2/A1) = 1/13 x 3/51
Rolling two 6s followed by one 3 on three tosses of a fair die.
The individual events are independent. The probability of the combined event is (1/6) x (1/6) x (1/6) = 1/216
independent events
the occurence of one event does not affect the probability of the other event occurring. if A and B have individual probabilities P(A) and P(B), the probability that A and B occur together is P(A and B)= P(A) x P(B)
dependent events
the outcome of one event affects the outcome of another event the probability that dependent events A and B occur together is P (A and B) = P (A) x P (B given A) = P(A) x P (B/A) , where P (B given A) is the probability of event B given the occurrence of event A
non overlapping events
they cannot occur together, like the outcome of a coin toss (heads or tails) the probability for non-overlapping events A and B, P(A or B) = P(A) + P(B)
At least how many times do you have to roll a fair die to be sure that the probability of rolling at least one 2 is greater than 8 in 10 (0.80)?
1 - (5/6) ^ 9 9 rolls ( test all exponents until the decimal is greater than 0.8)
Suppose you roll a die 4 times. What is the probability of getting at least one six?
1 - 1/6 = 5/6 (5/6)^4 = 625/1296 1 - 625/1296 = 671/1296
what is the probability of drawing at least 1 ace when you draw a card from a standard deck 6 times, replacing the card each time
1-P (no A)^6 = 1 - (12/13)^6 = 1 - 0.619 = 0.381
overlapping events
if they can occur together, like the outcome of picking a queen or a club the probability that either A or B occurs: P (A or B) = P(A) + P(B) - P(A and B)
Purchasing 5 winning lottery tickets in a row when each ticket has a 1 in 5 chance of being a winner
(1/5)^ 5 = 1/3125
Randomly selecting a four-person committee consisting entirely of Americans from a pool of 12 Americans and 16 Canadians.
12/28 x 11/27 x 10/26 x 9/25 = 0.0242
what is the probability that in a standard shuffled deck of cards you will draw a 5 or a spade
P(5 or spade) = P(5) + P(spade) - P(5 and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13
Getting a green light at a busy intersection at least once in six times through the intersection, given that the light in your direction is green 4/10 of the time
P(A) = 0.4 P(not A) = 1 - 0.4 = 0.6 n = 6 1 - (0.6)^6 = 0.953
The probability of drawing an ace or a spade from a deck of cards is the same as the probability of drawing the ace of spades.
The statement does not make sense because there is one card that is the ace of spades but more than one card that is either an ace or a spade.
The probability of getting heads and tails when you toss a coin is 0, but the probability of getting heads or tails is 1.
The statement makes sense because heads and tails are the only possible outcomes and it is impossible to get both heads and tails on a single coin toss.
My chance of getting a 5 on the roll of one die is 1/6, so my chance of getting at least one 5 when I roll three dice is 3/6.
This does not make sense because the real probability would be 1 - (5/6) ^3 , which is not equal to 3 divided by 6.
In which of the following situations are the events overlapping? Select all that apply.
We want to know whether a person selected at random is a Democrat or a man.
In which of the following situations would we be interested in an either/or probability? Select all that apply.
We want to know whether a person selected at random is a Democrat or a man. We roll a die, hoping for a 2 or a 5.
Being dealt three jacks off the top of a standard deck of well-shuffled cards.
dependent and probability ( without replacement ) P(A and B and C) = P(A) x P(B given A) x P(C given A and B) 1/13 x 1/17 x 1/25 = 1/5525
Drawing either a spade or a club from a regular deck of cards
non overlapping 13/52 + 13/52 = 1/2
Drawing either a black eight or a red three on one draw from a regular deck of cards
non overlapping, probability is 2/52 + 2/52 = 1/26 + 1/26 = 1/13
At least once rule (independent events)
suppose probability of an event A occurring in one trial is P(A). if all trials are independent, the probability that event A occurs at least once in n trials is: P (at least 1 event) = 1 - P (no event A)