Question

a second jar of candy contains only sweet pieces, 16 of which are yellow and 11 are blue. charlotte selects two pieces of candy from this new jar at random, without replacement. determine the probabilities of:
(a) (i) both pieces being blue.
(ii) both pieces being the same colour.
(iii) the second piece being yellow, given the first piece was blue.
(b) if charlotte didnt like the yellow flavour and kept selecting (and removing) pieces at random until she selected the one blue piece she wanted, calculate the probability of 5 pieces being selected in total.

applicants to a special division in the australian military forces used to undertake a fitness test with five components: sprint, endurance run, push-ups, pull-ups and squats. their fitness level on each component is classified as either \at standard\ or \below standard\. to pass the test, applicants must perform at an at standard fitness level in each of the five components. in increase intake this year, it was decided that applicants were allowed to omit any one of the five components to increase their chances to pass the test.
the table below shows the number of at standard and below standard performances in each component.

fitness level	component of the fitness test
	sprint	endurance run	push-ups	pull-ups	squats	total
at standard	22	80	46	31	11	190
below standard	18	6	10	9	47	90
total	40	86	56	40	58	280

(a) a military officer assesses the results and chooses a performance at random. find the probability that this randomly chosen performance is:
(i) a performance in sprint;
(ii) a below standard performance in sprint;
(iii) a below standard performance, given that it is a performance in sprint.
(b) the military officer groups the performances by component and chooses three performances in squats. find the probability that all three are at standard fitness level.

Explanation:

First Problem (Candy Jar Probabilities)

Step1: Total candies in jar

Total = 18 + 11 = 29

Step2 (a)(i): Probability both blue

First blue: $\frac{11}{29}$, second blue (no replacement): $\frac{10}{28}$. Multiply:
$\frac{11}{29} \times \frac{10}{28} = \frac{110}{812} = \frac{55}{406}$

Step3 (a)(ii): Probability same colour

Case 1: Both yellow: $\frac{18}{29} \times \frac{17}{28} = \frac{306}{812}$
Case 2: Both blue: $\frac{55}{406} = \frac{110}{812}$
Sum: $\frac{306 + 110}{812} = \frac{416}{812} = \frac{104}{203}$

Step4 (a)(iii): Conditional prob yellow after blue

After 1 blue removed, total = 28, yellow = 18:
$\frac{18}{28} = \frac{9}{14}$

Step5 (b): Probability 5 pieces selected total

This means first 4 are yellow, 5th is blue:
First 4 yellow: $\frac{18}{29} \times \frac{17}{28} \times \frac{16}{27} \times \frac{15}{26}$
5th blue: $\frac{11}{25}$
Multiply:
$\frac{18 \times 17 \times 16 \times 15 \times 11}{29 \times 28 \times 27 \times 26 \times 25} = \frac{792720}{14250600} = \frac{26424}{475020} = \frac{2202}{39585} = \frac{734}{13195}$

Step1: Total performances = 200

Step2 (a)(i): Probability performance in Sprint

Total Sprint performances = 52:
$\frac{52}{200} = \frac{13}{50}$

Step3 (a)(ii): Below Standard in Sprint

Below Standard Sprint = 30:
$\frac{30}{200} = \frac{3}{20}$

Step4 (a)(iii): Conditional Below Standard | Sprint

Use conditional probability: $P(B|A) = \frac{P(A \cap B)}{P(A)}$
$\frac{30/200}{52/200} = \frac{30}{52} = \frac{15}{26}$

Step5 (b): All 3 Squats are At Standard

Total At Standard Squats = 11, total Squats = 56. No replacement:
First: $\frac{11}{56}$, second: $\frac{10}{55}$, third: $\frac{9}{54}$
Multiply:
$\frac{11}{56} \times \frac{10}{55} \times \frac{9}{54} = \frac{990}{166320} = \frac{1}{168}$

Answer:

(a)(i) $\boldsymbol{\frac{55}{406}}$
(a)(ii) $\boldsymbol{\frac{104}{203}}$
(a)(iii) $\boldsymbol{\frac{9}{14}}$
(b) $\boldsymbol{\frac{734}{13195}}$

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Second Problem (Military Fitness Test Probabilities)