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Number of selections

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Number of white balls

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Number of red balls

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Number of red balls selected

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Number of white balls selected

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Probability of a certain number of balls being red

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Probability that a white ball is drawn

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Probability of picking a red ball

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probability less than so many balls are white

"}}, "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "information", "variableReplacements": [], "steps": [{"minValue": "p", "scripts": {}, "correctAnswerFraction": false, "type": "numberentry", "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "marks": 1, "prompt": "

What is the probability that the ball is white?

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What is the probability that the ball is red? Please enter you answer as a fraction or as a decimal with 5 significant figures.

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A ball is chosen at random from the bag.

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Find the probability that $\\var{rballs}$ of the selected balls are red.

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Find the probability that $0$ balls are white.

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Find the probability that fewer than $\\var{wballs}$ of the selected balls are white.

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A ball is chosen at random from the bag, its colour noted, then returned to the bag. This is repeated $\\var{nballs}$ times.

Part (a)

\n

There are $\\var{x}$ white balls and $\\var{y}$ red balls. Thus, there are a total of $\\var{x}+\\var{y}=\\var{total_balls}$ balls.

\n

i) The probability of picking a white ball is $\\frac{Number White Balls}{TotalNumberBalls} = \\frac{\\var{x}}{\\var{total_balls}} = \\var{p}$.

\n

ii) The probability of picking a red ball is $\\frac{Number Red Balls}{TotalNumberBalls} = \\frac{\\var{y}}{\\var{total_balls}} = \\var{1-p}$

\n

\n

Part (b)

\n

As $\\var{nballs}$ balls are selected with replacement and there are only two colours of ball, we have a binomial distibution. Let $X$ be the number of white balls selected, then $X\\sim B(\\var{nballs},p)$, where $p=\\var{p}$. Similarly, let $Y$ be the number of red balls selected, then  $Y\\sim B(\\var{nballs},q)$, where $q=\\var{oneminusp}$.

\n

i) The probability that exactly $\\var{rballs}$ of the selected balls are red is thus given by:

\n

${\\var{nballs} \\choose \\var{rballs}} q^\\var{rballs}(1-q)^{\\var{nballs}-\\var{rballs}} = \\var{prob_n_red}$.

\n

ii) The probability that none of the selected balls are white is given by:

\n

${\\var{nballs} \\choose \\var{0}} p^\\var{0}(1-p)^{\\var{nballs}} = (1-p)^{\\var{nballs}} = \\var{prob_zero_white}$.

\n

iii) For this part we need to find the cumulative binomial probability, the probability that less than $\\var{wballs}$ ball are white is given by.

\n

$\\sum_{i=0}^{\\var{wballs}-1}{\\var{nballs} \\choose \\var{i}} p^\\var{i}(1-p)^{\\var{nballs}-i} = \\var{prob_white_less_than}$.

\n

\n

\n

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A bag contains $\\var{x}$ white balls and $\\var{y}$ red balls.

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This question is concerned with finding probabilities from a normal distribution.

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Variance

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mean

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standard deviation

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Part a

\n

Given a distribution $X\\sim N(\\mu,\\sigma^2)$, then if $Y=\\frac{X-\\mu}{\\sigma}$, $Y\\sim N(0,1)$.

\n

In this case, $\\mu = \\var{mu}$ and $\\sigma = \\var{sigma}$, hence $Y = \\simplify{{onesigma}X-{musigma}}$.

\n

\n

Part b

\n

The following parts show which areas under the normal curve need to be found.

\n

i) {drawintegral1(mu,sigma,ubound)}

\n

ii) {drawintegral2(mu,sigma,lbound1,ubound1)}

\n

iii) {drawintegral3(mu,sigma,lbound2)}

\n

iv) {drawintegral4(mu,sigma,lbound3,ubound3)}

If $X \\sim N(\\var{mu},\\var{variance})$, then  $Y\\sim N(0,1)$ if

\n

$Y =$[[0]]$X+$Gap 1

\n

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$P(-\\infty \\leq X \\leq \\var{ubound})$

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$P(\\var{lbound1} \\leq X \\leq \\var{ubound1})$

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$P(\\var{lbound2} \\leq X \\leq \\infty)$

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$P(\\{-\\infty\\leq X \\leq \\var{ubound3}\\} \\cup \\{\\var{lbound3} \\leq X \\leq \\infty\\})$

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Using a calculator, or the statistical tables, find the following probabilities:

\n

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