Determine p from given mean or variance

Use the relationship between mean, variance, and p to find the probability parameter p (e.g., given Var(X) = 20, find p).

7 questions

CAIE FP2 2018 June Q9
9 At a ski resort, the probability of snow on any particular day is constant and equal to \(p\). The skiing season begins on 1 November. The random variable \(X\) denotes the day of the skiing season on which the first snowfall occurs. (For example, if the first snowfall is on 5 November, then \(X = 5\).) The variance of \(X\) is \(\frac { 4 } { 9 }\).
  1. Show that \(4 p ^ { 2 } + 9 p - 9 = 0\) and hence find the value of \(p\).
  2. Find the probability that the first snowfall will be on 3 November.
  3. Find the probability that the first snowfall will not be before 4 November.
  4. Find the least integer \(N\) so that the probability of the first snowfall being on or before the \(N\) th day of November is more than 0.999 .
CAIE FP2 2018 June Q7
7 The probability that a driver passes an advanced driving test has a fixed value \(p\) for each attempt. A driver keeps taking the test until he passes. The random variable \(X\) denotes the number of attempts required for the driver to pass. The variance of \(X\) is 3.75 .
  1. Show that \(15 p ^ { 2 } + 4 p - 4 = 0\) and hence find the value of \(p\).
  2. Find \(\mathrm { P } ( X = 5 )\).
  3. Find \(\mathrm { P } ( 3 \leqslant X \leqslant 7 )\).
CAIE FP2 2015 November Q6
6 A biased coin is tossed repeatedly until a head is obtained. The random variable \(X\) denotes the number of tosses required for a head to be obtained. The mean of \(X\) is equal to twice the variance of \(X\). Show that the probability that a head is obtained when the coin is tossed once is \(\frac { 2 } { 3 }\). Find
  1. \(\mathrm { P } ( X = 4 )\),
  2. \(\mathrm { P } ( X > 4 )\),
  3. the least integer \(N\) such that \(\mathrm { P } ( X \leqslant N ) > 0.999\).
CAIE FP2 2018 November Q8
8 Lan starts a new job on Monday. He will catch the bus to work every day from Monday to Friday inclusive. The probability that he will get a seat on the bus has the constant value \(p\). The random variable \(X\) denotes the number of days that Lan will catch the bus until he is able to get a seat. The probability that Lan will not get a seat on the Monday, Tuesday, Wednesday or Thursday of his first week is 0.4096 .
  1. Show that \(p = 0.2\).
  2. Find the probability that Lan first gets a seat on Monday of the second week in his new job.
  3. Find the least integer \(N\) such that \(\mathrm { P } ( X \leqslant N ) > 0.9\), and identify the day and the week that corresponds to this value of \(N\).
CAIE FP2 2017 Specimen Q6
6 A biased coin is tossed repeatedly until a head is obtained. The random variable \(X\) denotes the number of tosses required for a head to be obtained. The mean of \(X\) is equal to twice the variance of \(X\).
  1. Show that the probability that a head is obtained when the coin is tossed once is \(\frac { 2 } { 3 }\).
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  2. Find \(\mathrm { P } ( X = 4 )\).
  3. Find \(\mathrm { P } ( X > 4 )\).
  4. Find the least integer \(N\) such that \(\mathrm { P } ( X \leqslant N ) > 0.999\).
OCR Further Statistics AS 2023 June Q7
7 A town council is planning to introduce a new set of parking regulations. An interviewer contacts randomly chosen people in the town and asks them whether they are in favour of the proposal. The first person who is not in favour of the regulation is the \(R\) th person interviewed. It can be assumed that the probability that any randomly chosen person is not in favour of the proposal is a constant \(p\), and that \(p\) does not equal 0 or 1 . Assume first that \(\mathrm { E } ( R ) = 10\).
  1. Determine \(\mathrm { P } ( R \geqslant 14 )\). Now, without the assumption that \(\mathrm { E } ( R ) = 10\), consider a general value of \(p\).
    It is given that \(\mathrm { P } ( R = 3 ) - 0.4 \times \mathrm { P } ( R = 2 ) - a \times \mathrm { P } ( R = 1 ) = 0\), where \(a\) is a positive constant.
  2. Determine the range of possible values of \(a\).
OCR Further Statistics 2021 June Q4
7 marks
\(\mathbf { 4 }\) & \(\mathbf { ( a ) }\) & Geometric & M1 & 1.1 & Stated explicitly
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QuestionAnswerMarkAOGuidance
Mean \(= 400 \div 100 ( = 4 )\) and \(p = 1 /\) meanM12.4Use mean (or P(1) etc) to deduce \(p\) ("Determine", so justification needed for 0.25)Needs to deduce \(p\) in part (a), not defer it to (b)
\multirow{3}{*}{4}\multirow[t]{3}{*}{(b)}Probability is \(0.75 ^ { 6 } ( = 0.1779785 \ldots )\)M13.3SC Geo(0.2): \(0.8 { } ^ { 6 }\) M1A0
Or: 0.177978 or 0.177979 or better seen, or \(1 - [ \mathrm { P } ( 1 ) + \ldots + \mathrm { P } ( 6 ) ]\) with evidence, e.g. formulaM1Allow ± 1 term
Expected frequency \(=\) probability \(\times 100 = 17.798\)A1 [2]2.117.798 correctly obtained, with sufficient evidence, www\(100 - \Sigma\) (other frequencies): SC B1
\multirow{6}{*}{4}\multirow{6}{*}{(c)}Ho: data consistent with (geometric)B11.1
Both, allow equivalents, but not "evidence that ...". 9.005 or 9.01
Compare their \(\Sigma X ^ { 2 }\) with 11.07
E.g. \(\mathrm { H } _ { 0 } : X \sim \operatorname { Geo } ( p )\) Allow Geo(0.25)
\(\Sigma X ^ { 2 } = 9.005\)B11.1
\(9.005 < 11.07 ( v = 5 )\)B11.1
Do not reject \(\mathrm { H } _ { 0 }\).M1ft1.1Correct first conclusion, ft on their 9.005 or on 12.59, needs like-with-likeAllow from comparison with 12.59 but nothing else
Insufficient evidence that a geometric distribution is not a good fit.A1ft [5]2.2bContextualised, not too definite (needs double negative) Don't penalise "Geo(0.25)"Allow addition slip in \(\Sigma X ^ { 2 }\) SC Geo(0.2): can get full marks if given data used, \(\Sigma X ^ { 2 } = 4.54\) used gets B1B1B0M1A1