What is the probability of two events happening? My friend asked me,\u00a0\u201cI need a clear day with no rain on either a Saturday or Sunday to plant my garden. I don\u2019t care which.\u00a0There is typically a 30% chance of rain on any given day during the summer months.\u201d<\/p>\n
Since the probability of two events both happening is the product of each, 0.30 times 0.30 equals 0.09. I explained, \"The probability that it will rain both days is 9%. Therefore,\u00a0the probability you will have at least one nice day is 91%.\u201d<\/p>\n
Note that since probabilities are fractions, multiplying them makes answers smaller.\u00a0 Both events occurring is less likely than only one of them occurring.\u00a0The answer satisfied my friend, but it was a little oversimplified.<\/p>\n
The correct statement should have been. \u201cThe probability of two\u00a0independent<\/em>\u00a0events both happening is the product of each.\u201d\u00a0My knowledge of meteorology is a little lacking here, but I suspect the probability it rains on the second day is affected by whether or not it rained the first day.<\/p>\n For example, water laying on the ground after the first rain may be evaporated to add to humidity the second day. Or golfers who get rained out on Saturday may curse the rain gods who reply with more rain on Sunday.<\/p>\n We can look at this situation in a different light using a basic deck of playing cards.\u00a0From a deck of 52 face cards, what's the probability of drawing two cards that are of the same suit?<\/p>\n You'll first want to ask the question, did you put the first card back into the deck before drawing the second (with or without replacement)?\u00a0If you did put the first card back into the deck, the second draw is independent of the first. The probability the second draw matches the first is 13\/52 or 1\/4\u2014a 25% probability. (There are 13 cards of each suit in a deck.)<\/p>\n But if you are still holding the first card, there are now only 12 cards of that suit and only 51 cards in the deck.\u00a0 So the probability is 12\/51 which is about 0.235 or just slightly less than 25%.<\/p>\n Here is a slightly more complicated question.\u00a0 What is the probability of drawing hearts twice?\u00a0 The probability that the first card is a heart is 1\/4.\u00a0 With replacement, the second card also has a 1\/4 probability of being a heart, so the product is 1\/4 x 1\/4 = 1\/16 or 0.0625.<\/p>\n Without replacement, (you are still holding the first card) the probability is 1\/4 x 12\/51 = 0.059.<\/p>\n Notice the wording of the two questions:<\/p>\n A similar question\u2014after you draw the first card (without replacement), what\u2019s the probability the second card is a heart?\u00a0 If the first is a heart, the second probability is 12\/51.\u00a0 If the first is not a heart, the second probability is 13\/51.<\/p>\n A friend told me she has 10 grandchildren, all of whom are girls. \u201cWhat\u2019s the probability of that?\u201d she asked.\u00a0 Did she mean the probability of all girls, or all the same?\u00a0 I think the more logical question would be \u201call the same\u201d because she would have been just as amazed had there been 10 boys.<\/p>\n Around the world about 49% of babies at birth are girls and 51% are boys. This makes for very interesting demographic studies, and differences can be found based on characteristics of mothers such as age, race, health, and economic status.<\/p>\n Let's assume the birth rates are both 50%.\u00a0 We will also assume that the gender selection of each child is independent, which may or may not be true, based on the technique of getting babies started.\u00a0 The probability of 10 girls is 1\/2 x 1\/2 ten times or (0.5)10 = 1\/1024 = 0.00098.<\/p>\n Now it can be 10 girls or 10 boys.\u00a0 When the first baby is born, he\/she sets the standard for the rest.\u00a0 The other 9 must match the first so there are 9 trials.\u00a0 (Perhaps \u201ctrial\u201d is not the right word in deference to those mothers having the babies.)\u00a0 The probability is (0.5)9 = 1\/512 = 0.00195.<\/p>\n Consider the two characteristics \u201cfemale\u201d and \u201cleft-handed.\u201d Roughly 50% of humans are female and about 10% are left-handed. (Assume gender and hand preference are independent.)\u00a0 Here are four questions that can be asked.<\/p>\n 1) A left-handed female (ie. both left-handed\u00a0and<\/em>\u00a0female)?\u00a0 0.50 x 0.10 = 0.05.\u00a0 The restrictions that both conditions must be met means the probability decreases.<\/p>\n 2) Either left-handed\u00a0or<\/em>\u00a0female?\u00a0 Here it\u2019s the possibilities that are increased rather than the restrictions.\u00a0 The number of humans that meet either of the conditions is larger.\u00a0 We will add probabilities.<\/p>\n However, if we consider all left-handed persons and then all females, the left-handed females get counted twice.\u00a0So the probability is as follows:<\/p>\n 3) Either left-handed or female, but not both?\u00a0 We must subtract the probability of being both (Question 1) again.\u00a0 0.55 \u2013 0.05 = 0.50. This just happened to come out to be 50%, but\u00a0it cannot be generalized.<\/p>\n 4) Neither left-handed nor female?\u00a0 This is the opposite of Question 2 above.\u00a0 A person who is neither left-handed nor female, has to be a right-handed male.<\/p>\n Since 90% of humans are right-handed, the probability is 0.90 x 0.50 = 0.45.\u00a0 Of all humans 45% are right-handed males, and 55% are not, either being left-handed or a female (or both).\u00a0 Of course that adds up to 100% or 1 if expressed as a decimal.<\/p>\n Let\u2019s return to the Saturday and Sunday weather question which is quite similar to the gender\/preferred-hand question.\u00a0 For simplicity's sake, let's assume the probability of rain is 30% for both days, and rain the second day is independent of the first. The probability of it not raining is 0.70 each day.<\/p>\n (Rain, rain)\u00a0 0.30 x 0.30 = 0.09<\/p>\n That means either Saturday or Sunday, but not both?\u00a0 Either (rain, no rain) 0.30 x 0.70 = 0.21 or (no rain, rain) 0.7 x 0.30 = 0.21.<\/p>\n Since this is an increase of possibilities, (an \u201cor\u201d question) we add the two.\u00a0 0.21 + 0.21 = 0.42. (There is no overlap where something was counted twice.)<\/p>\n Or we can use the method of Question 3 above: the sum of each, subtract the product twice.\u00a0 0.30 + 0.30 \u2013 2(0.09) = 0.42.<\/p>\n Same answer. (I love it when math is supposed to work and it does.)<\/p>\n All the possible results for the two days are (rain, rain), (rain, no rain), (no rain, rain), (no rain, no rain).\u00a0 Of the four, the last three contain a no-rain day. \u00a00.30 x 0.70 + 0.70 x 0.30 + 0.70 x 0.70 = 0.21 x 0.21 x 0.49 = 0.91.<\/p>\n Note that Questions 1 and 3 are opposites:<\/p>\n Often it is easier to calculate the probability of what you don\u2019t want and subtract it from 1.\u00a0 The opposite of at-least-one-good day is rain-both days.\u00a0 1 \u2013 0.09 = 0.91.<\/p>\n The probability that both events happen is the product of each if they're independent. If they're not, the probability of the second must be modified based on the results of the first.\u00a0The probability that either one or the other happens is the sum of their probabilities, less the product of both if they overlap.\u00a0 It may be easier to calculate 1 \u2013 the opposite of the desired probability.<\/p>\nPlaying Cards<\/h2>\n
Drawing the Same Cards Multiple Times<\/h3>\n
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Predicting Gender in Children<\/h2>\n
What is the probability of 10 babies\u00a0being\u00a0all girls?<\/h3>\n
What is the probability of 10 babies\u00a0being\u00a0all the same gender?<\/h3>\n
Increased Restrictions or Possibilities<\/h2>\n
What is the probability of a randomly selected human?<\/h3>\n
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Two Day\u2019s Weather<\/h2>\n
What's the probability that it rains both days?<\/h3>\n
What's the probability that it rains one day<\/h3>\n
What's the probability that there's one day with no rain?<\/h3>\n
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Conclusion<\/h2>\n