(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine. [Hint : There ar...
Question 1 Report
(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine.
[Hint : There are 4 fives and 2 red nines in a pack of 52 cards]
(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If /PQ/ = 5m and /QR/ = 3m, find the bearing of R from P, correct to the nearest degree.
Answer Details
(a) A pack has 4 fives and 2 red nines. Drawing a five and drawing a red nine are mutually exclusive events (a five cannot also be a red nine), so the probabilities add.
\(P(\text{five or red nine}) = \frac{4}{52} + \frac{2}{52} = \frac{6}{52} = \dfrac{3}{26}\)
(b) Take P as origin and measure position as (East, North). A bearing \(\theta\) gives the direction \((\sin\theta, \cos\theta)\).
- \(Q = 5(\sin150°,\ \cos150°) = (2.500,\ -4.330)\)
- \(\vec{QR} = 3(\sin060°,\ \cos060°) = (2.598,\ 1.500)\)
- \(R = (2.500+2.598,\ -4.330+1.500) = (5.098,\ -2.830)\)
Relative to P, R lies to the East and to the South, so the bearing is between 090° and 180°.
\(\text{Bearing} = 180° - \tan^{-1}\!\left(\dfrac{5.098}{2.830}\right) = 180° - 60.96° = 119.04°\)
Bearing of R from P \(\approx \mathbf{119°}\) (to the nearest degree).
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