NCERT Solutions Class 10 Maths Chapter 14 Probability

Attach the right word to the right number. Examiners use this fill-in style to test definitions, so each answer has to be one precise value or term and never a full sentence. Here is how to read each blank correctly.

• Fixed pairing: the value zero always carries the words impossible event, and the value one always carries the words sure or certain event, so write the number and its word together.
• Why two ones: parts (i) and (iv) both come out as one but for different reasons. In (i) the event and its opposite split every outcome into two groups that do not overlap, while in (iv) the elementary events simply list every single outcome of the experiment exactly once.
• The bound in (v): no probability can ever fall below zero or rise above one, so any stray value such as a negative number or a number bigger than one is ruled out at once.
• Scoring tip: writing the term beside its number, for instance zero with impossible, earns both the value mark and the definition mark, which is exactly how the board splits the credit on this question.

(i) 1; (ii) 0, impossible; (iii) 1, sure/certain; (iv) 1; (v) 0 and 1.

Give a one-line reason, not just yes or no. The board awards the mark for the explanation, so each verdict needs a short cause written beside it.

• Part (i): the result turns on the car's mechanical condition. A faulty car may fail to start many times in a row, so starting and not starting clearly carry different chances and are not equally likely.
• Part (ii): the result turns on the player's skill. A trained player scores far more often than a beginner misses, so the two outcomes are unequal.
• Part (iii): a guessed true or false answer is simply right or wrong, exactly one of the two is correct, and a blind guess favours neither, so the two outcomes really are equally likely.
• Part (iv): the standard Class 10 assumption is biological symmetry between a boy and a girl, so the two are taken as equally likely.

Stating the reason in a few words next to each verdict is what separates a full-mark answer from a bare guess.

(iii), (iv) equally likely (blind guess / biological symmetry); (i), (ii) not, because they depend on machine condition and player skill.

Anchor `fair' to equal probability. The marker looks for the idea that both teams hold a half chance, so always write the two equal probabilities out.

• Symmetry: a fair coin is physically balanced, so there is no reason for it to land head up more often than tail up, and that balance is exactly what makes the two outcomes equally likely.
• Equal chance: since head and tail are equally likely, each team, having picked one face before the toss, has the very same chance of winning the decision, with neither side carrying any advantage.
• Why a coin: because no team can claim the rule favoured the other, a neutral coin toss rather than a referee's personal choice is the accepted way to decide who starts with the ball.

Because P(head) = P(tail) = 12, both teams are equally likely to win, so the toss is unbiased and fair.

Scan for the value outside the range. In an MCQ like this you need not evaluate every option fully, only find the one that breaks the bound.

• Spot the negative: a probability is a count of favourable outcomes over a positive total, so it can never come out below zero, which means any negative option is wrong on sight.
• The answer: option (B) is a clearly negative number, so it cannot be a probability and is the choice you want without further work.
• Check the rest: the other three all sit safely between zero and one, two of them obviously and the percentage once you read it as a decimal, so none of them breaks the rule.

Spotting the single out-of-range value is faster than testing each option, and it is the exact skill the question is testing.

(B) -1.5, since it is negative and every probability lies in [0,1].

A one-step use of the complement rule. The board sets this kind of question only to confirm you know that an event and its opposite add to one.

• The idea: since the two probabilities must add to a whole, the chance that the event does not happen is simply whatever is left after you remove the chance that it does.
• The work: here only a small part of the total belongs to the event, so the rest of it belongs to the opposite event, and subtracting gives the answer at once.
• Sanity check: the answer lies between zero and one and the two chances add back to one exactly, so both the value and the method are sound; show the rule before the arithmetic to secure the method mark.

0.95.

Read the word `only' carefully. That single word tells you the bag is pure lemon, and that fact alone fixes both answers at once.

• Orange: there is not a single orange candy, so the favourable count for orange is zero, and zero over any positive total is zero, which makes drawing an orange candy an impossible event.
• Lemon: the bag is entirely lemon, so whichever candy comes out is certain to be lemon, making it a sure event with probability one.
• Neat link: the two answers also illustrate complements, since they add to one, so not getting orange is exactly the same as getting lemon.

Orange is impossible (0); lemon is certain (1).

Recognise the complement first. The phrase about not having the same birthday is the direct opposite of having the same birthday, so the two probabilities must total one.

• What is given: you are handed the large chance for the not-same case, and the question asks only for the small leftover, so a single subtraction from one finishes the work.
• Why it is small: the answer comes out tiny, which makes sense because two students chosen at random sharing a birthday is genuinely a rare event.
• Check: the given chance and your answer add back to one exactly, which confirms that the pair really is complementary and that the arithmetic is right.

0.008.

Fix the total first, then count favourables. The most common slip is forgetting that the denominator is the total number of balls, not just the red ones.

• Total: there are eight balls in all, so eight is the denominator for every part, and writing this sum out plainly earns the first method mark even before any fraction appears.
• Red: the red probability is simply the red count over that total of eight.
• Not red: you may either subtract the red chance from one or count the black balls directly, and both routes land on the same answer, which is a reassuring built-in check.
• Final check: the red and not-red answers add to one, confirming the complement was handled correctly.

Red =38, not red =58.

Keep one common denominator throughout. Every probability here sits over the same total, so once you fix that denominator the rest is just reading off the colour counts.

• Direct colours: the red and white probabilities are read straight off as their counts over the common total, with nothing to work out beyond the simple fractions.
• Not green: the fastest route in the exam is to add the two non-green colours and put that count over the total, which lands on the answer at once and agrees with subtracting the green chance from one.
• Final check: the three colour probabilities together add up to one, which confirms that not a single marble was miscounted along the way.

Red 517, white 817, not green 1317.

Add the counts, then simplify carefully. The total is the sum of all four coin counts, and every probability in this question carries that same denominator.

• The 50p coin: its favourable count is the number of fifty paise coins over the grand total, and that fraction cancels neatly to a much simpler form once you divide top and bottom.
• Not a five rupee coin: the neat exam route is to add up all the coins that are not five rupee ones and place that count over the total, which matches subtracting the five rupee chance from one.
• Reduce fully: the marking scheme expects both answers in lowest terms, so finishing the cancellation is part of earning the full credit here, not an optional flourish.

59 for a 50p coin; 1718 for not a  5 coin.

A clean single-step ratio question. The board adds a picture only to make the problem friendly; the maths is just favourable over total.

• The count: there are thirteen fish in the tank and five of them are male, so the male probability is the male count over the total, and it is already in lowest terms.
• Quick check: the female chance would be the other count over the same total, and the two together add to one, which confirms that the total of thirteen is right.
• For full marks: write the total as the sum of male and female fish before forming the fraction, so the reasoning is explicit on the page.

513.

Recognise the sure event in part (iv). Three of the parts are ordinary counts, but the last part quietly checks whether you can spot a certain event.

• Pointing at eight: only one sector on the spinner shows the number eight, so just one outcome out of the eight is favourable, and that gives the smallest of the answers here.
• An odd number: four of the eight numbers are odd, so the favourable count is four out of eight, which reduces to one half once you cancel the common factor.
• Greater than two: six of the numbers lie above two, leaving out only the one and the two themselves, so six out of eight reduces to three quarters.
• Less than nine: the largest label printed on this spinner is eight, so every single outcome is already less than nine, which makes the event certain and its probability one.

Seeing that the phrase less than nine quietly covers the whole spinner is the intended trap, and naming it as a sure event is what shows full understanding to the examiner.

18, 12, 34, 1 for parts (i)-(iv).

Read each phrase precisely. All three answers turn out to be a half, but for different reasons, so the listing of numbers matters more than the value.

• Primes: the primes on a die are two, three and five, because one has only a single factor and four and six are composite, which leaves exactly three favourable faces.
• Between two and six: the word between is exclusive, so it picks only three, four and five and leaves out the two endpoints, again giving three faces.
• Odd numbers: the odd faces are one, three and five, a third set of three out of the six faces.

Each set of three out of six reduces to a half. The lesson is that identical-looking answers can hide different reasoning, so always write the qualifying numbers down before counting them.

All three probabilities equal 12.

Build the deck picture once, then answer all six. Most card questions reduce to counting how many cards match a description out of the fifty two.

• Red kings: there are exactly two of them, one in each red suit, so the favourable count is two out of the whole deck.
• Face cards: every suit has a king, a queen and a jack, so across the four suits there are twelve face cards in all.
• Red face cards: this is the one part needing real care, because a card must be both red and a face card at the same time, so you take only the six that belong to the two red suits, not all twelve.
• Named single cards: the jack of hearts and the queen of diamonds are unique, so each is just one card out of the deck.
• Spades: the spades form one complete suit of thirteen, which reduces neatly to a quarter of the deck.

Reducing each fraction fully to its lowest terms is exactly what the marking scheme rewards, so do not leave any answer half cancelled.

126, 313, 326, 152, 14, 152.

Track how the pile changes between draws. The whole question is about noticing that removing the queen alters both the total and the cards that are left.

• First draw: there are five cards with a single queen among them, so the queen has a chance of one in five on the opening pick.
• After removal: once that queen is set aside, only four cards remain and the queen is no longer one of them, so always redraw the reduced pile before working out the second chance.
• Second draw, ace: the ace is still sitting in the pile, so its chance is one out of the four cards that are left.
• Second draw, queen: the queen is now gone, so the favourable count is zero and drawing a queen has become an impossible event.

The clever part is the last case, where students who forget the queen was removed wrongly write one quarter instead of zero.

15; then 14 for an ace and 0 for a queen.

Form the fraction, then reduce fully. The marking scheme wants the final answer in lowest terms, so the cancellation step is itself part of the credit.

• The fraction: the total is every pen in the lot, good and defective together, and the good pens over that total give the raw fraction before any simplifying.
• Reduce it: the top and bottom share a common factor, and dividing both through by it turns the messy fraction into a clean one in lowest terms.
• Check: the defective chance is the leftover, and the good chance plus the defective chance add back to one, which confirms the good-pen answer is right.

1112.

Re-count the lot before the second draw. The question is really testing the idea of drawing without replacement, so the second part hangs on what was removed.

• Part one: four of the twenty bulbs are defective, so the defective chance is four over twenty, which cancels to one fifth.
• Part two: a good bulb has already left the lot, so the good bulbs are now down to fifteen and the total down to nineteen, giving the new chance over nineteen.
• The key point: because a good bulb was taken out, both the favourable count and the total drop by one; had a defective bulb left instead, the good count would have stayed put, so always note which kind was removed.

15 then 1519.

Each part is a clean counting exercise. The total of ninety stays fixed throughout, and only the favourable count changes from one part to the next.

• Two-digit numbers: the quickest route is to remove the nine single-digit numbers from the ninety discs, which leaves the rest as two-digit, so the favourable count is whatever is left after that simple subtraction.
• Perfect squares: list the squares one by one and stop at the last one that does not exceed ninety, since the next square already overshoots it, and that listing gives you exactly nine favourable values.
• Multiples of five: you need only divide ninety by five to get how many multiples there are, and that count over the total gives the chance without any listing at all.

Writing the favourable count clearly before you reduce keeps each answer tidy and earns the method mark even when the final fraction is simple.

910, 110, 15.

Watch the repeated letter. The whole point of this question is that one letter sits on two faces, which doubles its chance.

• The denominator: the die still has six equally likely faces, so six is always the denominator no matter which letter you are asked about.
• Letter A: this letter occupies two of the six faces, so its chance is two over six, which is double that of any letter sitting on a single face.
• Letter D: this one sits on just a single face, so its chance is the plain one over six.
• Common slip: counting the five distinct letters and writing one fifth for A miscounts the sample space, because it is the faces, not the letter names, that are equally likely.

P(A) = 13 and P(D) = 16.

Keep pi as the fraction twenty-two sevenths. The board answer for this question is usually left as a fraction, so avoid switching to a rounded decimal too early in the working.

• New idea here: this is geometric probability, where outcomes are points in a region, so the chance is the area of the favourable part over the area of the whole region, not a count of outcomes.
• Favourable area: the circle has a radius of half a metre, and putting that radius into the area of a circle, with pi kept as a fraction, gives its area as a tidy fraction of a square metre rather than a decimal.
• Total area: the whole region is the rectangle, and its area is simply its length times its breadth in square metres.
• Divide: dividing the circle area by the rectangle area gives the probability as a single fraction, which is the form the board prefers.

Two checks make the answer safe: the circle clearly fits inside the rectangle so the chance cannot exceed one, and the value sits comfortably between zero and one as every probability must.

1184 (about 0.13).

Use the complement to save work. Once you find one probability, the other is just one minus it, because buying and not buying cover every possible case.

• She buys: buying means the pen is good, so subtract the defective pens from the total to get the good count, and that count over the total gives the buying chance.
• She does not buy: not buying means the pen is defective, and you can reach this either by counting the defective pens directly or by taking the buying chance away from one.
• Check and reduce: both routes to the not-buying chance agree, the two answers add back to one as a built-in check, and reducing each fraction fully is needed for full marks.

Buys 3136, does not buy 536.

Use the rise-and-fall pattern of the pair counts. The number of pairs climbs steadily to a peak at the middle sum and then falls back in mirror image, which is a fast way to fill the whole table without listing every pair by hand.

• Reading the table: the pair counts start at one for the smallest sum, rise by one each step up to the peak at the middle sum, then drop back by one each step to one again at the largest sum, so the table is symmetric about the centre.
• Fixed denominator: every one of those counts sits over the same total of thirty six, and because the counts themselves add up to thirty six, the probabilities across the row add to one as they must.
• Why the claim fails: the student confuses the number of distinct sums with the number of equally likely outcomes, but the genuinely equally likely outcomes are the thirty six ordered pairs, not the eleven sums.
• The consequence: since each sum inherits a different chance from how many pairs build it, giving every sum the same value cannot be right, because the middle sums clearly arise far more often than the extreme sums at either end.

Probabilities follow the 1 to 6 to 1 pattern over 36; the 111 claim is wrong because the sums are not equally likely.

Build the eight-outcome sample space once. Three coin tosses always give eight equally likely results, and writing them all out removes any guesswork about the counts.

• When Hanif wins: he wins only when all three tosses match, which happens for the all-heads result and the all-tails result, so just two of the eight outcomes are winning ones.
• When Hanif loses: losing covers all the remaining mixed results, which is six of the eight outcomes, and you can reach this either by counting them or by subtracting the winning chance from one.
• The check: the winning and losing chances add back to one, which confirms that the two events are complementary and that no outcome was accidentally left out.

34.

Set up the pair sample space, then take the complement. The phrase at least once is a classic signal to work out the opposite event first.

• Sample space: throwing a die twice gives thirty six equally likely ordered pairs, and that fixed total is the denominator for both parts of the question.
• No five at all: for this case each throw must avoid the five, which leaves five safe numbers per throw, so multiplying the safe choices for the two throws gives the favourable count of pairs.
• At least one five: this is simply everything else, so subtract the no-five chance from one rather than trying to add up the separate cases.
• Two checks: the two answers add to one, and a direct count of the pairs that contain a five, taking care not to count the double five twice, confirms the at-least-one answer independently.

No 5 either time 2536; at least one 5 1136.

Decide the equally likely sample space before judging the claim. Both parts stand or fall on whether the stated outcomes are genuinely equally likely.

• Part one is wrong: the trap is folding the two mixed results into a single one-of-each outcome, which makes the three groups unequal; the true equally likely list is the four ordered pairs, so one of each really carries half, not a third.
• Part two is correct: the two groups odd and even each contain exactly three of the six faces, so they are equally likely and the chance of an odd number comes out as one half just as the argument claims.
• The lesson: the two parts look like the same reasoning, but it is valid only when the underlying outcomes are balanced, which is why one argument holds and the other fails.

(i) Not correct; (ii) correct.