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IB Mathematics: Applications and Interpretation HL · 13.1: Probability · 13.1.7
Independent events Problem set A One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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A shop's card reader is out of order on 0.45 of days and its till printer is out of order on 0.6 of days. The two faults are independent. Find the probability that both are out of order on the same day.
- Independent, so \(P(A\cap B)=P(A)\times P(B)\)
- \(0.45\times 0.6=0.27\)
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For events A and B, \(P(A)=0.5\), \(P(B)=0.3\) and \(P(A\cup B)=0.65\). Determine whether A and B are independent. Justify your answer.
- \(P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.3-0.65=0.15\)
- \(P(A)\times P(B)=0.5\times 0.3=0.15\)
- The two are equal, so the events are independent.
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A website's search bar and its login page fail independently. The probability that the search bar fails is 0.4. The probability that both fail is 0.14. Find the probability that the login page fails.
- Independent, so \(P(A\cap B)=P(A)\times P(B)\)
- \(0.14=0.4\times P(B)\)
- \(P(B)=0.14\div 0.4=0.35\)
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Events A and B are independent, with \(P(A)=\frac{3}{8}\) and \(P(A\cap B)=\frac{3}{20}\). Find \(P(B)\).
- Independent, so \(P(A\cap B)=P(A)\times P(B)\)
- \(\frac{3}{20}=\frac{3}{8}\times P(B)\)
- \(P(B)=\frac{3}{20}\div \frac{3}{8}=\frac{2}{5}\)
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Events A and B have P(A) = 0.5 and P(B) = 0.2.
(a) A and B are independent. Find P(A ∪ B).
- Independent: \(P(A\cap B)=0.5\times0.2=0.1\)
- \(P(A\cup B)=0.5+0.2-0.1=0.6\)
(b) State, with a reason, whether A and B could also be mutually exclusive.
- Mutually exclusive needs \(P(A\cap B)=0\).
- But \(P(A\cap B)=0.1\), so they cannot be both.
In context
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In a board game, two fair six-sided dice are rolled and the score is the product of the two numbers.
(a) Use a sample space diagram to find the probability that the score is at least 15.
- Draw a 6 by 6 grid with the first die down the side and the second across the top. Each cell holds the product.
- 13 of the 36 products are at least 15.
- \(P=\frac{13}{36}\)
(b) The game is played 144 times. How many times would you expect the score to be at least 15?
- Expected number = probability × number of trials
- \(\frac{13}{36}\times 144=52\)
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A fair four-sided die (numbered 1 to 4) and a fair eight-sided die (numbered 1 to 8) are rolled and the scores are added.
(a) Find the probability that the total is 6.
- Grid of totals: 4 rows (the 4-sided die) and 8 columns (the 8-sided die), 32 outcomes in all.
- 4 of them fit.
- \(P=\frac{4}{32}=\frac{1}{8}\)
(b) Find the probability that the total is a multiple of 4.
- Grid of totals: 4 rows (the 4-sided die) and 8 columns (the 8-sided die), 32 outcomes in all.
- 8 of them fit.
- \(P=\frac{8}{32}=\frac{1}{4}\)
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A sports club has 120 members. Of these, 45 swim, 50 cycle and 15 do both.
(a) Draw a Venn diagram to show this information.
- Start in the middle: 15 do both.
- Only swim: 45 − 15 = 30. Only cycle: 50 − 15 = 35.
- Neither: 120 − (30 + 15 + 35) = 40
(b) Find the probability that a person chosen at random does neither.
- \(P=\frac{40}{120}=\frac{1}{3}\)
(c) Find the probability that a member swims, given that they do not cycle.
- Not cycle means outside the cycle circle: 30 + 40 = 70 people.
- \(P=\frac{30}{70}=\frac{3}{7}\)
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One card is chosen at random from a standard pack of 52 cards. H is the event "the card is a heart" and K is the event "the card is a king".
(a) Are H and K mutually exclusive? Give a reason.
- The king of hearts is a heart and a king, so \(P(H\cap K)=\frac{1}{52}\neq0\).
(b) Are H and K independent? Show your working.
- \(P(H)=\frac{13}{52}=\frac14\), \(P(K)=\frac{4}{52}=\frac1{13}\)
- \(P(H)\times P(K)=\frac1{52}=P(H\cap K)\), so H and K are independent.
Stretch
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Two events are independent, with \(P(A)=0.3\) and \(P(B)=0.5\). Find \(P(A\cup B)\).
- Independent, so \(P(A\cap B)=0.3\times 0.5=0.15\)
- \(P(A\cup B)=0.3+0.5-0.15=0.65\)
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Events A and B are independent, with \(P(A)=0.6\) and \(P(A\cup B)=0.82\). Find \(P(B)\).
- Independent: \(P(A\cap B)=P(A)P(B)\)
- \(P(A\cup B)=P(A)+P(B)-P(A)P(B)\), so \(0.82=0.6+P(B)(1-0.6)\)
- \(P(B)=\frac{0.82-0.6}{0.4}=0.55\)
jmaths.xyz · AI HL · 13.1.7 Independent events