### Video Transcript

A card is selected at random from a pack of 52 cards which are numbered one to
52. What is the probability that the selected card has a number which is neither a
perfect square nor a multiple of six?

In this question, we have a pack of 52 cards numbered one to 52. We need to calculate the probability that a card selected at random is neither a
perfect square nor a multiple of six.

The perfect squares between one and 52 are one, four, nine, 16, 25, 36, and 49. There are seven such cards. The multiples of six between one and 52 are six, 12, 18, 24, 30, 36, 42, and 48. There are eight such cards. Note that 36 appears in both lines. As such, there are 14 cards that are a perfect square or a multiple of six.

We recall that probability can be written as a fraction, where the numerator is the
number of favorable outcomes and the denominator is the total number of possible
outcomes. Note that this holds when every item is equally likely to be selected, as in this
case. Since there are 52 cards in total and 14 are perfect squares or multiples of six, we
have 52 minus 14 favorable outcomes. This is equal to 38. The required probability is therefore equal to 38 out of 52.

Since both the numerator and denominator are even, we can divide them both by
two. As such, our fraction simplifies to 19 over 26. Hence, the probability of selecting a card at random which is neither a perfect
square nor a multiple of six is 19 out of 26.