Miscellaneous Exercise answers: Probability

Open NCERT p. 310Matches NCERT’s answer

(i) all will be blue?

1. Total marbles = 10 + 20 + 30 = 60. 5 marbles are drawn together.
2. Total ways to draw 5 marbles = C(60, 5).
3. Ways to draw 5 blue marbles = C(20, 5), since there are 20 blue marbles.
4. P(all blue) = C(20,5) / C(60,5) = 34/11977.

Answer34/11977

(ii) atleast one will be green?

1. "At least one green" is the complement of "no green at all".
2. No green means all 5 come from the 10 red + 20 blue = 30 non-green marbles.
3. P(no green) = C(30,5) / C(60,5).
4. P(at least one green) = 1 − C(30,5)/C(60,5) = 4367/4484.

Answer4367/4484

Watch this explained “Three questions in disguise”, 14:47 into Knowing one probability hands you its opposite for free

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1. Total ways to draw 4 cards from 52 = C(52, 4).
2. Ways to choose 3 diamonds out of 13 diamonds = C(13, 3).
3. Ways to choose 1 spade out of 13 spades = C(13, 1).
4. Favourable ways = C(13,3) × C(13,1).
5. P = [C(13,3) × C(13,1)] / C(52,4) = 286/20825.

Answer286/20825

Watch this explained “Seven cards from fifty-two”, 11:06 into Assuming the outcomes are equally likely recovers the counting rule

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(i) P(2)

1. 6 faces in all: 2 show '1', 3 show '2', 1 shows '3'.
2. P(2) = 3/6 = 1/2.

Answer1/2

(ii) P(1 or 3)

1. Getting '1' and getting '3' cannot happen together on one roll, so add their probabilities.
2. P(1) = 2/6, P(3) = 1/6.
3. P(1 or 3) = 2/6 + 1/6 = 3/6 = 1/2.

Answer1/2

(iii) P(not 3)

1. P(not 3) = 1 − P(3) = 1 − 1/6 = 5/6.

Answer5/6

Watch this explained “A die with repeated faces”, 14:07 into Assuming the outcomes are equally likely recovers the counting rule

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(a) one ticket

1. 10,000 tickets, 10 winning tickets, so 9990 are non-winning.
2. P(no prize with 1 ticket) = 9990/10000 = 999/1000.

Answer999/1000

(b) two tickets

1. Choosing 2 tickets out of the 9990 non-winning ones, out of all ways to choose 2 tickets from 10,000.
2. P(no prize with 2 tickets) = C(9990,2) / C(10000,2) = 1108779/1111000.

Answer1108779/1111000

(c) 10 tickets

1. Choosing 10 tickets out of the 9990 non-winning ones, out of all ways to choose 10 tickets from 10,000.
2. P(no prize with 10 tickets) = C(9990,10) / C(10000,10).

Answer339164885989139627914993428516/342576807891019297980996468625

Watch this explained “Seven cards from fifty-two”, 11:06 into Assuming the outcomes are equally likely recovers the counting rule

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(a) you both enter the same section?

1. The 100 places are 40 in one section and 60 in the other, filled at random.
2. The two places that you and your friend get are equally likely to be any 2 of the 100 places, so the number of equally likely pairs is C(100, 2) = 4950.
3. Both in the section of 40: C(40, 2) = 780 pairs. Both in the section of 60: C(60, 2) = 1770 pairs.
4. P(same section) = (780 + 1770)/4950 = 2550/4950 = 17/33.

Answer17/33

(b) you both enter the different sections?

1. 'Different sections' is the complement of 'same section'.
2. P(different sections) = 1 − 17/33 = 16/33.
3. Check: one from each section can be chosen in 40 × 60 = 2400 ways, and 2400/4950 = 16/33.

Answer16/33

Watch this explained “Two groups in a class”, 13:59 into Knowing one probability hands you its opposite for free

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1. There are 3! = 6 equally likely ways to put the 3 letters into the 3 envelopes.
2. Count how many of the 6 ways put every letter in the wrong envelope (no match at all): 2 ways.
3. "At least one letter correct" is the complement of "no letter correct".
4. P(at least one correct) = 1 − 2/6 = 4/6 = 2/3.

Answer2/3

Watch this explained “Three letters, three envelopes”, 12:44 into Knowing one probability hands you its opposite for free

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(i) P(A ∪ B)

1. Use the addition rule: P(A∪B) = P(A) + P(B) − P(A∩B).
2. P(A∪B) = 0.54 + 0.69 − 0.35 = 0.88.

Answer22/25

(ii) P(A′ ∩ B′)

1. A′∩B′ is the complement of A∪B (De Morgan's law).
2. P(A′∩B′) = 1 − 0.88 = 0.12.

Answer3/25

(iii) P(A ∩ B′)

1. A∩B′ is the part of A that is outside B.
2. P(A∩B′) = P(A) − P(A∩B) = 0.54 − 0.35 = 0.19.

Answer19/100

(iv) P(B ∩ A′)

1. B∩A′ is the part of B that is outside A.
2. P(B∩A′) = P(B) − P(A∩B) = 0.69 − 0.35 = 0.34.

Answer17/50

Watch this explained “Two candidates, one entrance test”, 12:50 into Adding two probabilities double-counts the overlap, so the overlap comes back off

Open NCERT p. 311Matches NCERT’s answer

1. List who is male, or over 35, or both: Harish (M), Rohan (M), Salim (M, also over 35), Sheetal (F, over 35).
2. Alis is female and 28, so she is the only one who is neither.
3. 4 of the 5 persons satisfy "male or over 35".
4. P = 4/5.

Answer4/5

Watch this explained “Where the overlap has to be spotted”, 18:12 into Adding two probabilities double-counts the overlap, so the overlap comes back off

Open NCERT p. 311Matches NCERT’s answer

(i) the digits are repeated?

1. For a 4-digit number above 5000 made from 0, 1, 3, 5, 7, the first digit must be 5 or 7.
2. With repetition allowed there are 2 × 5 × 5 × 5 = 250 such arrangements. One of them is 5000 itself, which is not greater than 5000, so there are 250 − 1 = 249 numbers.
3. Divisible by 5 means the last digit is 0 or 5: 2 × 5 × 5 × 2 = 100 arrangements. Again 5000 is one of them, so 100 − 1 = 99 numbers.
4. P = 99/249 = 33/83.
5. (If 5000 were wrongly counted, you would get 100/250 = 2/5. The question says greater than 5000, so it must be left out.)

Answer33/83

(ii) the repetition of digits is not allowed?

1. Without repetition the first digit is 5 or 7, and the other three places take different digits from the four left: 2 × 4 × 3 × 2 = 48 numbers, all above 5000.
2. The last digit must be 0 or 5. Ending in 0: 5 _ _ 0 and 7 _ _ 0 each fill the middle from 3 digits in 3 × 2 = 6 ways, so 12 numbers.
3. Ending in 5: the first digit must be 7 (5 is already used), 7 _ _ 5 gives 3 × 2 = 6 numbers.
4. Favourable = 12 + 6 = 18, so P = 18/48 = 3/8.

Answer3/8

Watch this explained “The rule comes back”, 7:40 into Assuming the outcomes are equally likely recovers the counting rule

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1. Each wheel shows one of 10 digits, and no digit repeats across the 4 wheels.
2. Number of possible 4-digit sequences with no repeats = 10 × 9 × 8 × 7 = 5040.
3. Only one of these sequences opens the lock.
4. P(right sequence) = 1/5040.

Answer1/5040

Watch this explained “The denominator counts outcomes”, 12:13 into Assuming the outcomes are equally likely recovers the counting rule