(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.
(b) The table below shows the distribution of 20 cards labelled A - E.
(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?
(b) The table below shows the distribution of 20 cards labelled A - E.
| Card | A | B | C | D | E |
| Frequency | 3 | 4 | 7 | 5 | 1 |
(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?
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Correct Answer: Option n
Explanation:
(a)
\(< BAD = 106°\) (Given)
\(/AD/ = /AB/\) (given)
\(\therefore < ADC = \frac{180° - 106°}{2} = 37°\) (base angles of an isosceles triangle)
(b) Total card = 20
(i) P(card picked is E) = \(\frac{1}{20}\)
(ii) P(B and A or C or D or E) or P(A or C or D or E and B)
= \(\frac{4}{20} \times \frac{3 + 7 + 5 + 1}{19} + \frac{3 + 7 + 5 + 1}{20} \times \frac{4}{19}\)
= \(\frac{64}{380} + \frac{64}{380} = \frac{32}{95}\)
(a)
\(< BAD = 106°\) (Given)
\(/AD/ = /AB/\) (given)
\(\therefore < ADC = \frac{180° - 106°}{2} = 37°\) (base angles of an isosceles triangle)
(b) Total card = 20
(i) P(card picked is E) = \(\frac{1}{20}\)
(ii) P(B and A or C or D or E) or P(A or C or D or E and B)
= \(\frac{4}{20} \times \frac{3 + 7 + 5 + 1}{19} + \frac{3 + 7 + 5 + 1}{20} \times \frac{4}{19}\)
= \(\frac{64}{380} + \frac{64}{380} = \frac{32}{95}\)