Probability Equations

The three probabilities must add to one.

1. The probabilities of all possible outcomes add to 1.
2. So x+0.25+0.35=1.
3. Hence x=0.40=\frac{2}{5}.

Use the fact that the outcomes are exhaustive.

1. Add the probabilities: 2k+3k+5k=1.
2. This gives 10k=1, so k=\frac{1}{10}.

Multiply the head probabilities, then use the valid probability range.

1. Independence gives p^2=0.49.
2. The algebraic roots are \pm0.7.
3. A probability cannot be negative, so p=0.7=\frac{7}{10}.

Express the total number of counters in terms of the unknown.

1. The total number of counters is x+4.
2. Form \frac{4}{x+4}=\frac{2}{5}. Cross-multiplying gives 20=2x+8.
3. So 2x=12 and x=6.

Include the constant when adding the probability expressions.

1. The probabilities sum to 1, so x+2x+x+0.2=1.
2. Thus 4x=0.8 and x=0.2=\frac{1}{5}.
3. The probabilities are 0.2, 0.4 and 0.4, all within the valid range.

Use expected frequency to form a multiplication equation.

1. Expected frequency equals number of trials multiplied by probability.
2. So 200p=60, giving p=\frac{60}{200}=\frac{3}{10}.

Use ratio parts for the original counts and compare the final counts.

1. Write the original counts as 3k red and 5k blue.
2. A red probability of \frac{1}{2} means the final counts of the two colours are equal.
3. So 3k+8=5k, giving k=4.
4. The original number of blue counters is 5\times4=20.

The union counts the overlap only once.

1. Let the intersection probability be x.
2. The addition rule gives 0.55+0.4-x=0.7.
3. So x=0.25=\frac{1}{4}.

Form an equation for both independent alarms sounding.

1. Multiply probabilities for independent events: 0.4q=0.14.
2. So q=0.14\div0.4=0.35=\frac{7}{20}.

Use the complement of at least one success.

1. Neither attempt succeeds with probability 1-0.91=0.09.
2. Thus (1-p)^2=0.09. Since 1-p is a probability, it cannot be negative, so 1-p=0.3.
3. Therefore p=0.7=\frac{7}{10}.

Convert the losing probability to a winning probability first.

1. The winning probability is 1-0.85=0.15.
2. Let the total number be N. Then \frac{12}{N}=0.15.
3. So N=12\div0.15=80.

Find how many counters are not red, and what probability they make up.

1. There are 2\times21=42 blue counters, so 42+21=63 counters are not red.
2. The probability of not red is 1-0.3=0.7, so these 63 counters are 0.7 of all the counters.
3. If there are N counters in total, 0.7N=63, so N=63\div0.7=90.
4. The number of red counters is 90-63=27 (check: \frac{27}{90}=0.3).

Adding white balls changes both the favourable count and the total.

1. Let n be the number added. Then there are n+1 white balls out of n+4 balls.
2. Form \frac{n+1}{n+4}=\frac{1}{2}.
3. This gives 2n+2=n+4, so n=2 white balls were added.

Weight each conditional damage probability by its route probability.

1. A tree diagram has two paths to "damaged": by air and damaged, or not by air and damaged.
2. By air and damaged: 0.3\times x=0.3x. Not by air and damaged: 0.7\times0.5=0.35.
3. Follow the two routes to a damaged parcel.

4. Add the paths and set equal to the total: 0.3x+0.35=0.38.
5. So 0.3x=0.03 and x=0.1.

The unknown weights the two source probabilities; do not average them equally.

1. The two paths to black give \frac{p}{4}+\frac{3(1-p)}{4}=\frac{2}{5}.
2. Weight each bag by its selection probability.

3. Multiply by 20: 5p+15(1-p)=8.
4. Simplify to 15-10p=8, so p=\frac{7}{10}.

Let N be the total number of cards, and remember there is one card fewer for the second pick.

1. Let N be the total number of cards. Without replacement, the probability that both are gold is \frac{7}{N}\times\frac{6}{N-1}=\frac{42}{N(N-1)}.
2. Set this equal to \frac{1}{5}: \frac{42}{N(N-1)}=\frac{1}{5}, so N(N-1)=210.
3. Expand and rearrange: N^2-N-210=0.
4. Factorise: (N-15)(N+14)=0, so N=15 (a number of cards cannot be -14).
5. There are 15 cards, so the number of silver cards is 15-7=8.

Include both orders for exactly one success, then apply the stated range.

1. Exactly one success means success then failure, or failure then success. Each has probability p(1-p).
2. Include both orders for exactly one success.

3. So 2p(1-p)=\frac{12}{25}, which expands to 2p-2p^2=\frac{12}{25}.
4. Multiply by 25: 50p-50p^2=12. Rearrange and divide by 2: 25p^2-25p+6=0.
5. Factorise: (5p-2)(5p-3)=0, so p=\frac{2}{5} or p=\frac{3}{5}.
6. The question says p<\frac{1}{2}, so p=\frac{2}{5}.

Count both colour orders and keep every root that makes physical sense.

1. There are n+4 counters. Different colours means blue then red, or red then blue.
2. Blue then red: \frac{4}{n+4}\times\frac{n}{n+3}. Red then blue: \frac{n}{n+4}\times\frac{4}{n+3}. Together: \frac{8n}{(n+4)(n+3)}.
3. Track the changing bag total in both colour orders.

4. Set equal to \frac{8}{15} and divide both sides by 8: \frac{n}{(n+4)(n+3)}=\frac{1}{15}, so 15n=(n+4)(n+3).
5. Expand: 15n=n^2+7n+12, so n^2-8n+12=0.
6. Factorise: (n-2)(n-6)=0, so n=2 or n=6. Both are whole numbers of counters, so both are possible.

Form a conditional ratio from the morning row, then recover the overall total.

1. The morning total is (x+6)+(2x-4)=3x+2.
2. The conditional probability gives \frac{x+6}{3x+2}=\frac{3}{5}.
3. Cross-multiply: 5x+30=9x+6, giving x=6.
4. There are 20 morning tickets and 18 evening tickets, so the total is 38.

Separate the cases according to the colour transferred before forming the equation.

1. If blue is transferred, B has 3 blue counters out of n+3. If red is transferred, B has 2 blue counters out of n+3.
2. Track the two possible transfers into box B.

3. The total probability equation is \frac{3}{5}\times\frac{3}{n+3}+\frac{2}{5}\times\frac{2}{n+3}=\frac{13}{40}.
4. Combine the two terms to give \frac{13}{5(n+3)}=\frac{13}{40}.
5. Cross-multiply and cancel 13: 40=5(n+3).
6. Hence n+3=8 and n=5.