Class 11 Maths NCERT Solutions Chapter 14 Miscellaneous Exercise – Probability

The box has 10+20+30=60 marbles in total, and 5 are drawn without replacement, so the sample space has \binom{60}{5} equally likely outcomes.

(i) All 5 blue means all 5 are chosen from the 20 blue marbles:

P(\text{all blue})=\dfrac{\binom{20}{5}}{\binom{60}{5}}=\dfrac{34}{11977}

(ii) 'At least one green' is easier via the complement: 'no green' means all 5 are chosen from the 10+20=30 non-green marbles.

P(\text{no green})=\dfrac{\binom{30}{5}}{\binom{60}{5}}

P(\text{at least one green})=1-P(\text{no green})=1-\dfrac{\binom{30}{5}}{\binom{60}{5}}=\dfrac{4367}{4484}

Choosing 4 cards from 52 gives a sample space of \binom{52}{4} equally likely outcomes.

3 diamonds can be chosen from the 13 diamonds in \binom{13}{3} ways, and 1 spade can be chosen from the 13 spades in \binom{13}{1} ways:

P(3\text{ diamonds, }1\text{ spade})=\dfrac{\binom{13}{3}\binom{13}{1}}{\binom{52}{4}}=\dfrac{286\times13}{270725}=\dfrac{286}{20825}

The die has 6 faces: 2 show '1', 3 show '2', and 1 shows '3'. So P(1)=\dfrac{2}{6}=\dfrac{1}{3}, P(2)=\dfrac{3}{6}=\dfrac{1}{2}, P(3)=\dfrac{1}{6}.

(i) P(2)=\dfrac{1}{2}

(ii) Since faces showing 1 and faces showing 3 can't both turn up at once, P(1\text{ or }3)=P(1)+P(3)=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{1}{2}

(iii) P(\text{not }3)=1-P(3)=1-\dfrac{1}{6}=\dfrac{5}{6}

There are 10,000 tickets and 10 winning tickets, so 9,990 tickets do not win.

(a) Buying 1 ticket: the probability it is a non-winning ticket is:

P(\text{no prize})=\dfrac{9990}{10000}=\dfrac{999}{1000}

(b) Buying 2 tickets: the sample space is all ways of choosing 2 tickets from 10,000, and 'no prize' means both chosen tickets come from the 9,990 non-winning tickets:

P(\text{no prize})=\dfrac{\binom{9990}{2}}{\binom{10000}{2}}=\dfrac{1108779}{1111000}\approx0.9980

(c) Buying 10 tickets: similarly, 'no prize' means all 10 chosen tickets come from the 9,990 non-winning tickets:

P(\text{no prize})=\dfrac{\binom{9990}{10}}{\binom{10000}{10}}\approx0.9900

Think of the 100 students being randomly split into a section of 40 and a section of 60. Once you know your own section, the probability your friend joins you there depends on how many spots are left in that section out of the remaining 99 students.

(a) If you land in the 40-section (probability \frac{40}{100}), your friend joins you with probability \frac{39}{99} (39 spots left out of 99 remaining students). Similarly for the 60-section, with probability \frac{60}{100}\times\frac{59}{99}.

P(\text{same section})=\dfrac{40}{100}\times\dfrac{39}{99}+\dfrac{60}{100}\times\dfrac{59}{99}=\dfrac{40\times39+60\times59}{100\times99}=\dfrac{1560+3540}{9900}=\dfrac{5100}{9900}=\dfrac{17}{33}

(b) 'Different sections' is the complement of 'same section':

P(\text{different sections})=1-\dfrac{17}{33}=\dfrac{16}{33}

There are 3!=6 ways to place 3 letters into 3 envelopes, all equally likely.

'At least one letter correctly placed' is easiest via the complement: 'no letter correctly placed' is a derangement of 3 objects. The number of derangements of 3 objects is D_3=2 (for objects 1, 2, 3 in envelopes 1, 2, 3, the only derangements are the two ways to cyclically shift all three).

P(\text{no letter correct})=\dfrac{D_3}{3!}=\dfrac{2}{6}=\dfrac{1}{3}

P(\text{at least one letter correct})=1-\dfrac{1}{3}=\dfrac{2}{3}

(i) Using the addition rule:

P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.54+0.69-0.35=0.88

(ii) By De Morgan's law, A'\cap B'=(A\cup B)':

P(A'\cap B')=1-P(A\cup B)=1-0.88=0.12

(iii) A\cap B' is 'A but not B', which is A with the overlap A\cap B removed:

P(A\cap B')=P(A)-P(A\cap B)=0.54-0.35=0.19

(iv) Similarly, B\cap A' is 'B but not A':

P(B\cap A')=P(B)-P(A\cap B)=0.69-0.35=0.34

Let M = 'male' = {Harish, Rohan, Salim}, so P(M)=\dfrac{3}{5}. Let O = 'over 35 years' = {Sheetal (46), Salim (41)}, so P(O)=\dfrac{2}{5}.

The overlap M∩O = 'male and over 35' = {Salim}, so P(M\cap O)=\dfrac{1}{5}.

Using the addition rule:

P(M\cup O)=P(M)+P(O)-P(M\cap O)=\dfrac{3}{5}+\dfrac{2}{5}-\dfrac{1}{5}=\dfrac{4}{5}

For a number to exceed 5000, the thousands digit must be 5 or 7 (digits 0, 1, 3 are all too small).

Thousands digit: 2 choices (5 or 7). Hundreds and tens digits: 5 choices each (any of the 5 digits, with repetition). This gives 2\times5\times5=50 combinations for the first three digits, and the units digit has 5 choices, for 50\times5=250 total numbers — but the single case 5-0-0-0 gives exactly 5000, which is not greater than 5000, so it must be excluded, leaving 250-1=249 valid numbers.

For divisibility by 5, the units digit must be 0 or 5 (2 choices). With the thousands digit fixed (2 choices) and the middle two digits free (5\times5=25 choices), this gives 2\times25\times2=100 combinations — again excluding the single invalid case of exactly 5000, leaving 100-1=99 valid numbers.

P(\text{divisible by }5)=\dfrac{99}{249}=\dfrac{33}{83}

Thousands digit: 2 choices (5 or 7). The remaining 3 positions are filled with 3 of the remaining 4 digits, in order: P(4,3)=4\times3\times2=24 ways. Total: 2\times24=48 numbers (no boundary case to exclude here, since digits can't repeat, so exactly 5000 is impossible to form).

For divisibility by 5, the units digit must be 0 or 5:

If thousands digit = 5: units digit can only be 0 (5 is already used). The middle two digits come from the remaining 3 digits {1,3,7}: P(3,2)=6 ways.

If thousands digit = 7: units digit can be 0 or 5 (2 choices). For each, the middle two digits come from the remaining 3 digits: P(3,2)=6 ways each, giving 2\times6=12.

Total divisible by 5: 6+12=18.

P(\text{divisible by }5)=\dfrac{18}{48}=\dfrac{3}{8}

Since no digit repeats, the number of possible 4-digit sequences from 10 digits is P(10,4)=10\times9\times8\times7=5040.

Only 1 of these sequences opens the lock, so:

P(\text{correct sequence})=\dfrac{1}{5040}