Briony takes a standard pack of 52 cards and throws some cards away. However, she makes sure she keeps all four aces among the remaining cards. She then selects four cards at random from these remaining cards. If the probability of her selecting the four aces is 1 0 0 1 1 , how many cards did she throw away?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Why can't we throw away 6 3 cards so that we have a negative probability of − 1 1 4 of drawing the first ace, − 1 2 3 of drawing the second ace, − 1 3 2 of drawing the third ace, and − 1 4 1 of drawing the fourth, so that the product is 1 0 0 1 1 ? Or am I just being nonsensical?
Log in to reply
The question stated that all four aces remained in the pack.
Problem Loading...
Note Loading...
Set Loading...
Assume that Briony has k cards left in the pack after she threw some cards away. Then, there are 2 4 k ( k − 1 ) ( k − 2 ) ( k − 3 ) different ways of selecting four cards. Since there is only one way of picking up four aces, the probability of such an event is k ( k − 1 ) ( k − 2 ) ( k − 3 ) 2 4 . Hence
k ( k − 1 ) ( k − 2 ) ( k − 3 ) 2 4 = 1 0 0 1 1
Therefore k ( k − 1 ) ( k − 2 ) ( k − 3 ) = 2 4 × 1 0 0 1 . There is only one value of k that satisfies this equation, and the fact that 1 0 0 1 = 7 × 1 1 × 1 3 helps to quickly find it: k = 1 4 . Thus the answer is 5 2 − 1 4 = 3 8 .