ProbabilityProbability — A Theoretical Approach
It is the inter-house cricket final, and the whole school is packed along the boundary rope. Riya, captain of Green House, and Aman, captain of Blue House, walk out to the pitch with the umpire. The umpire flicks a five-rupee coin into the air, and Riya calls heads.
"Not fair," Aman mutters. "That coin always lands her way."
Is a toss really fair? Nobody can say how one toss will land. But perhaps we can say something about what happens over many tosses.
You are the umpire. Toss the coin once, ten times, a hundred times, and watch the share of heads. Make at least 100 tosses in all.
Any single toss is a surprise, yet the share of heads keeps drifting back towards one half. That is what we mean when we call a coin fair or unbiased: it is perfectly symmetrical, so there is no reason for it to land on
Before the toss we already know the coin can land in only
A die works the same way. In this chapter a die always means a fair die. Throw it once and the possible outcomes are 1, 2, 3, 4, 5 and
Riya's Lucky Dip
Does every
"Only two colours," says Aman, handing over his ₹10. "Red or blue. So my chance is fifty-fifty."
Test Aman's claim. Draw a ball, note its colour and put it back — the bag shows which ball came out. Make at least 50 draws in all.
Aman's reasoning is broken. Red and blue are both possible outcomes, but they are not equally likely: there are four red balls and only one blue one, so red turns up far more often. A
Now look at the bag another way. Each of the five balls is just as likely to come out as any other, so the experiment does have five equally likely outcomes — and only one of them is blue. When the outcomes you care about are not equally likely, split them into smaller outcomes that are.
Note
Not every experiment has equally likely outcomes. From now on in this chapter, we assume that all the experiments have equally likely outcomes.
Counting Instead of Tossing
What you just did is the experimental idea of probability from Class IX. There, the
This only works when the experiment can be repeated a
Repeating is easy for a coin or a die. But some experiments are far too
Luckily, there is another way. For a fair coin we never really needed the hundred tosses: there are 2 equally likely outcomes and one of them is a head, so a head should turn up in about
Theoretical Probability
The assumption we need is that the outcomes are equally likely. With it, we get the definition at the heart of this chapter.
The
where we assume that the outcomes of the experiment are equally likely.
From now on, "probability" will mean theoretical probability. This definition was given by
The event builder. An event is any set of outcomes. Tap faces of the die to put them into your event E, and watch P(E) on the line from 0 to 1. Complete all three missions.
Here is why the formula works. The 6 equally likely faces of a die share the whole of the certainty between them, so each face carries a share of
From dice games to genetics
Probability theory began with games of chance. In the 16th century the Italian physician and mathematician
Let us put the definition to work on experiments with equally likely outcomes, starting with the umpire's coin.
Example 1 : Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
Solution : Tossing a coin once has two possible outcomes, head (H) and tail (T). Let E be the event "getting a head". Exactly one outcome, H, is favourable to E. So
In the same way, if F is the event "getting a tail", then P(F) = P(
Example 2 : A bag holds three balls of the same size: one red, one blue and one yellow. Kritika takes out one ball without looking into the bag. What is the probability that she takes out (i) the yellow ball? (ii) the red ball? (iii) the blue ball?
Solution : Kritika does not look, so each of the three balls is equally likely to come out. Let Y, R and B be the events "the ball is yellow", "the ball is red" and "the ball is blue".
The number of possible outcomes is
(i) Only
(ii) and (iii) In the same way, P(R) = and P(B) = .
Remarks
- An event that contains only one outcome of the experiment is called an
elementary event . In Example 1 the events E and F are elementary events; in Example 2 all three events Y, R and B are. - In Example 1, P(E) + P(F) = 1, and in Example 2, P(Y) + P(R) + P(B) = 1. This is always true: the probabilities of all the elementary events of an experiment add up to
.
Example 3 : A die is thrown once. What is the probability of getting (i) a number greater than 4? (ii) a number less than or equal to 4?
Solution : (i) Let E be the event "getting a number greater than 4". There are six possible outcomes, 1 to 6, and the outcomes favourable to E are the green faces 5 and 6. So
(ii) Let F be the event "getting a number less than or equal to 4". Its favourable outcomes are the purple faces 1, 2, 3 and 4, which is
Are E and F elementary events?
Two Events That Share Everything
Put Examples 1 and 3 side by side.
From Example 1: P(head) + P(tail) =
From Example 3: P(greater than 4) + P(at most 4) =
In both cases the second event is exactly "the first event does not happen": a number that is not greater than 4 is the same thing as a number less than or equal to 4. We write the event "not E" as . Between them, E and contain every outcome exactly once, so their probabilities always add up to 1.
For any event E,
The event , "not E", is called the complement of the event E. We say that E and are
Tip
When the outcomes you want are awkward to count but the ones you don't want are easy, count the easy ones: find and subtract it from 1. "At least one head in two tosses" has three favourable outcomes; its complement, "no head at all", has just one. You will use this shortcut again and again.
Speed Round: Count and Divide
Dice, bags of balls and complements. Count the favourable outcomes, count all the outcomes, divide and simplify. The working appears after each answer.
Get five right to move on.
Impossible and Sure
Two quick questions about a single throw of a die. (i) What is the probability of getting an 8? (ii) What is the probability of getting a number less than 7?
(i) A die has only
. An event that cannot happen is called an
(ii) Every face of the die shows a number less than 7, so every throw gives such a number: all
. An event that is certain to happen is called a
Sort these events into the right bucket.
The numerator of P(E) counts favourable outcomes, and there can never be fewer than none of them or more than all of them. So every probability lies between 0 and 1.
For every event E, . An impossible event has probability 0, and a sure event has probability 1.
A Deck of 52 Cards
Many board questions use a deck of playing cards, so let us get to know one. A deck has 52 cards in 4 suits of 13 cards each: spades, hearts, diamonds and clubs. Spades and clubs are black; hearts and diamonds are red. Each suit has an ace, a king, a queen, a jack and the numbered cards 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face cards.
Here is the whole deck, one suit to a row. Tap single cards, or tap a rank at the top of a column (or a suit at the start of a row) to select all of it at once. Complete all three missions.
Keep these counts handy: 52 cards, 26 red and
Example 4 : One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card will (i) be an ace, (ii) not be an ace.
Solution : Shuffling the deck well makes all 52 outcomes equally likely.
(i) Let E be the event "the card is an ace". There are
So P(E) =
(ii) Let F be the event "the card is not an ace". The cards that are not aces number
P(F) =
Remark
F is just , so the complement rule gives the answer in one line: P(F) = .
Speed Round: Card Sharp
One card from a well-shuffled deck, and a different event every time. Count the cards that fit; picture the deck grid if it helps.
Get five right to move on.
Example 5 : Sangeeta and Reshma play a tennis match. The probability that Sangeeta wins the match is 0.62. What is the probability that Reshma wins?
Solution : Let S and R be the events "Sangeeta wins" and "Reshma wins". A tennis match always has a winner, so exactly one of the two events happens: R is the complement of S, and the events S and R are
So
Example 6 : Savita and Hamida are friends. What is the probability that the two of them have (i) different birthdays? (ii) the same birthday? (Ignore leap years.)
Solution : Savita's birthday can be any day of the year. Whatever it is, Hamida's birthday can be any one of the
(i) For Hamida's birthday to differ from Savita's, it can fall on any day except Savita's birthday:
P(different birthdays) =
(ii) "The same birthday" is the complement of "different birthdays", so
P(same birthday) =
Example 7 : Class X of a school has 40 students: 25 girls and 15 boys. The class teacher has to choose one student as class representative. She writes each name on an identical card, puts all the cards in a bag, mixes them well and draws one card. What is the probability that the card shows the name of (i) a girl? (ii) a boy?
Solution : One card is drawn from 40 identical, well-mixed cards, so there are
(i)
(ii) 15 cards carry a boy's name, so P(boy) =
Note
"Boy" is the complement of "girl" here, so P(boy) = as well.
Example 8 : A box contains 3 blue, 2 white and 4 red marbles. One marble is drawn at random. What is the probability that it is (i) white? (ii) blue? (iii) red?
Solution : "At random" is a short way of saying that every marble is equally likely to be drawn. So the number of possible outcomes is
Let W, B and R be the events "the marble is white", "the marble is blue" and "the marble is red".
(i) 2 of the marbles are white, so P(W) =
(ii) P(B) =
Check: P(W) + P(B) + P(R) =
Design the Lucky Dip
The mela was a hit, and Riya is planning next year's stall. This time every bag will carry a label with an exact chance on it. Can you fill a bag that matches all three labels? Add and remove balls with the buttons, and the probabilities update as you go.
A bag with 3 red, 2 blue and 1 green ball matches all three labels at once: , and . Doubling every colour, to 6, 4 and 2, works just as well, because each probability depends on what
Example 9 : Harpreet tosses two different coins at the same time, a ₹1 coin and a ₹2 coin. What is the probability that she gets at least one head?
Before the solution, find the outcomes yourself. Each cell is one equally likely outcome: its row is the ₹1 coin and its column is the ₹2 coin.
Solution : Write H for head and T for tail. Two coins tossed together have four equally likely outcomes: (H, H), (H, T), (T, H) and (T, T). Here (H, T) means a head on the ₹1 coin and a tail on the ₹2 coin, which is a different outcome from (T, H).
The outcomes favourable to the event E, "at least
Note
The complement is quicker: is "no head", which has only the outcome (T, T). So P(E) = .
When Outcomes Cannot Be Counted
Every experiment so far had a finite number of outcomes: 2 for a coin, 6 for a die, 52 for a deck. But in some experiments the outcome can be any number between two given numbers, or any point inside a circle or a rectangle. There are infinitely many of those, so you cannot count them, and the definition of probability cannot be used as it stands. Here is the way out.
Example 10* : In a game of musical chairs, the person playing the music has been told to stop it at any moment within 2 minutes after starting. What is the probability that the music stops within the first half-minute?
* Not from the examination point of view.
Solution : The music can stop at any moment from 0 to 2 minutes, so the possible outcomes are all the points of the number line from 0 to 2.
Fig. 14.1 — every moment from 0 to 2 minutes is a possible outcome.
Let E be the event "the music stops within the first half-minute". Its favourable outcomes are the points of the stretch from 0 to ½. The whole stretch has length
Every moment is equally likely, so the probability is the favourable length as a fraction of the whole:
The same idea works with areas: the probability is the favourable area as a fraction of the total area.
Example 11* : A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig. 14.2. What is the probability that it crashed inside the lake shown in the figure?
* Not from the examination point of view.
Fig. 14.2 is live. The blue rectangle is the search region and the green rectangle is the lake. Drop search markers at random spots in the region, at least 200 in all, and watch what share of them lands in the lake.
Solution : The helicopter is equally likely to have crashed anywhere in the region.
Area of the lake
So P(the helicopter crashed in the lake) =
Example 12 : A carton holds 100 shirts, of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will accept only the good shirts, while Sujatha, another trader, will reject only the shirts with major defects. One shirt is drawn at random from the carton. What is the probability that it is (i) acceptable to Jimmy? (ii) acceptable to Sujatha?
Solution : One shirt is drawn at random from 100, so there are 100 equally likely outcomes. Here is the carton seen from above: 88 good shirts, 8 with minor defects and 4 with major defects.
(i) Jimmy accepts only the good shirts, so
P(acceptable to Jimmy) =
(ii) Sujatha accepts every shirt except the ones with major defects — the good shirts and the ones with minor defects. That is
P(acceptable to Sujatha) =
Two Dice, 36 Outcomes
Example 13 : Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8? (ii) 13? (iii) less than or equal to 12?
Solution : When the blue die shows 1, the grey die can show any of 1, 2, 3, 4, 5 and 6, and the same is true when the blue die shows 2, 3, 4, 5 or 6. So the outcomes fill the grid below (Fig. 14.3 in your textbook). In each ordered pair the first number is on the blue die (the row) and the second is on the grey die (the column).
Use the grid. Tap outcomes, or tap a die at the start of a row or the top of a column to select it all. Complete the three missions.
Notice that (1, 4) and (4, 1) are
(i) Let E be the event "the sum of the two numbers is 8". The favourable outcomes are the five pairs (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2), so P(E) = .
(ii) Let F be the event "the sum of the two numbers is 13". The largest possible sum is 6 + 6 = 12, so no outcome is favourable to F: P(F) =
(iii) Let G be the event "the sum of the two numbers is less than or equal to 12". Every one of the 36 outcomes is favourable, so P(G) =
Boss Round: Two Dice
The last challenge of the chapter: two dice, 36 outcomes and a new event every time. Picture the grid, count the favourable cells, divide by 36.
Get five right to finish the chapter's games.