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$A box contains 6 red cards, 5 blue cards and 9 yellow cards. The cards are identical except for the colours. Two cards are taken from the box without replacement. (a) Draw a tree diagram to show the probabilities of the possible outcomes. (b) Find, as a fraction in its simplest form, the probability that (i) the two cards are of the same colour, (ii) at least one of the cards is yellow. (c) A third card is taken from the box. Find, as a fraction in its simplest form, the probability that exactly two of the cards is blue. Firstcad Secend card$

$6$ . A box contains $6$ red cards, $5$ blue cards and $9$ yellow cards. $\newline$ The cards are identical except for the colours. $\newline$ Two cards are taken from the box without replacement. $\newline$ (a) Draw a tree diagram to show the probabilities of the possible outcomes. $\newline$ (b) Find, as a fraction in its simplest form, the probability that $\newline$ (i) the two cards are of the same colour, $\newline$ (ii) at least one of the cards is yellow. $\newline$ (c) A third card is taken from the box. $\newline$ Find, as a fraction in its simplest form, the probability that exactly two of the cards is blue. $\newline$ Firstcad $\newline$ Secend card

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Math Problems

Algebra 2

Find probabilities using the addition rule

Full solution

6

. A box contains

6

red cards,

5

blue cards and

9

yellow cards.

\newline

The cards are identical except for the colours.

\newline

Two cards are taken from the box without replacement.

\newline

(a) Draw a tree diagram to show the probabilities of the possible outcomes.

\newline

(b) Find, as a fraction in its simplest form, the probability that

\newline

(i) the two cards are of the same colour,

\newline

(ii) at least one of the cards is yellow.

\newline

\newline

Find, as a fraction in its simplest form, the probability that exactly two of the cards is blue.

\newline

Firstcad

\newline

Secend card

Total number of cards: Total number of cards = $6$ red + $5$ blue + $9$ yellow = $20$ cards.
Probability of two red cards: Probability of drawing a red card first = $\frac{6}{20}$ . Probability of drawing a red card second without replacement = $\frac{5}{19}$ . Probability of two red cards = $\left(\frac{6}{20}\right) * \left(\frac{5}{19}\right)$ .
Probability of two blue cards: Probability of drawing a blue card first = $\frac{5}{20}$ . Probability of drawing a blue card second without replacement = $\frac{4}{19}$ . Probability of two blue cards = $\left(\frac{5}{20}\right) * \left(\frac{4}{19}\right)$ .
Probability of two yellow cards: Probability of drawing a yellow card first = $\frac{9}{20}$ . Probability of drawing a yellow card second without replacement = $\frac{8}{19}$ . Probability of two yellow cards = $\left(\frac{9}{20}\right) * \left(\frac{8}{19}\right)$ .
Probability of two cards being the same color: Probability of two cards being the same color $= \text{Probability of two red cards} + \text{Probability of two blue cards} + \text{Probability of two yellow cards}.$
Probability of at least one yellow card: Probability of two cards being the same color = $(\frac{6}{20}) \times (\frac{5}{19}) + (\frac{5}{20}) \times (\frac{4}{19}) + (\frac{9}{20}) \times (\frac{8}{19})$ .
Probability of drawing a blue card third: Probability of at least one yellow card $=$ Probability of first yellow $+$ Probability of second yellow $-$ Probability of two yellow cards.
Probability that exactly two of the cards are blue: Probability of first yellow card = $\frac{9}{20}$ . Probability of second yellow card without replacement = $\frac{9}{19}$ (since the first card could be any color). Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} -$ Probability of two yellow cards.
Probability that exactly two of the cards are blue: Probability of first yellow card = $\frac{9}{20}$ . Probability of second yellow card without replacement = $\frac{9}{19}$ (since the first card could be any color). Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} -$ Probability of two yellow cards. Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} - \left(\frac{9}{20}\right) \times \left(\frac{8}{19}\right)$ .
Probability that exactly two of the cards are blue: Probability of first yellow card = $\frac{9}{20}$ . Probability of second yellow card without replacement = $\frac{9}{19}$ (since the first card could be any color). Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} -$ Probability of two yellow cards. Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} - \left(\frac{9}{20} \times \frac{8}{19}\right)$ . For part (c), after two cards are taken, there are $18$ cards left. Probability of drawing a blue card third = $\frac{5}{18}$ or $\frac{4}{18}$ depending on whether a blue card was drawn in the first two draws.
Probability that exactly two of the cards are blue: Probability of first yellow card = $\frac{9}{20}$ . Probability of second yellow card without replacement = $\frac{9}{19}$ (since the first card could be any color). Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} -$ Probability of two yellow cards. Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} - \left(\frac{9}{20}\right) \left(\frac{8}{19}\right)$ . For part (c), after two cards are taken, there are $18$ cards left. Probability of drawing a blue card third = $\frac{5}{18}$ or $\frac{4}{18}$ depending on whether a blue card was drawn in the first two draws. Probability that exactly two of the cards are blue = Probability of (blue, not blue, blue) + Probability of (not blue, blue, blue).
Probability that exactly two of the cards are blue: Probability of first yellow card = $\frac{9}{20}$ . Probability of second yellow card without replacement = $\frac{9}{19}$ (since the first card could be any color). Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} -$ Probability of two yellow cards. Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} - \left(\frac{9}{20}\right) \left(\frac{8}{19}\right)$ . For part (c), after two cards are taken, there are $18$ cards left. Probability of drawing a blue card third = $\frac{5}{18}$ or $\frac{4}{18}$ depending on whether a blue card was drawn in the first two draws. Probability that exactly two of the cards are blue = Probability of (blue, not blue, blue) + Probability of (not blue, blue, blue). Probability of (blue, not blue, blue) = $\left(\frac{5}{20}\right) \left(\frac{15}{19}\right) \left(\frac{4}{18}\right)$ . Probability of (not blue, blue, blue) = $\left(\frac{15}{20}\right) \left(\frac{5}{19}\right) \left(\frac{4}{18}\right)$ .
Probability that exactly two of the cards are blue: Probability of first yellow card = $\frac{9}{20}$ . Probability of second yellow card without replacement = $\frac{9}{19}$ (since the first card could be any color). Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} -$ Probability of two yellow cards. Probability of at least one yellow card = $\frac{9}{20} + \frac{9}{19} - \left(\frac{9}{20}\right) \left(\frac{8}{19}\right)$ . For part (c), after two cards are taken, there are $18$ cards left. Probability of drawing a blue card third = $\frac{5}{18}$ or $\frac{4}{18}$ depending on whether a blue card was drawn in the first two draws. Probability that exactly two of the cards are blue = Probability of (blue, not blue, blue) + Probability of (not blue, blue, blue). Probability of (blue, not blue, blue) = $\left(\frac{5}{20}\right) \left(\frac{15}{19}\right) \left(\frac{4}{18}\right)$ . Probability of (not blue, blue, blue) = $\left(\frac{15}{20}\right) \left(\frac{5}{19}\right) \left(\frac{4}{18}\right)$ . Probability that exactly two of the cards are blue = $\left(\frac{5}{20}\right) \left(\frac{15}{19}\right) \left(\frac{4}{18}\right) + \left(\frac{15}{20}\right) \left(\frac{5}{19}\right) \left(\frac{4}{18}\right)$ .