Jade has seven cards. Each card is labeled with a letter. A B C D E F G H J Jade picks one of her cards at random. Find the probability that the card she picks is a) labelled F, b) labelled with a letter in her name JADE c) labelled with a letter that has at least one line of symmetry

(a)

(b)

(c)

Step-by-step explanation:

Probability (P) of an event is the likelihood that the event will occur. It is given by;

P = number of favourable outcomes ÷ total number of events in the sample space.

Given letters of cards:

A B C D E F G H J

∴ Total number of events in sample space is actually the number of cards which is 7

If a card is picked at random;

(a) the probability P(F), that it is labelled F is given by;

P(F) = number of favourable outcomes ÷ total number of events in the sample space.

The number of favourable outcomes for picking an F = 1 since there is only one card labelled with F.

∴ P(F) = 1 ÷ 7

=> P(F) =

(b) the probability P(N), that it is labelled with a letter in her name JADE is given by;

P(N) = P(J) + P(A) + P(D) + P(E)

Where;

P(J) = Probability that it is labelled J

P(A) = Probability that it is labelled A

P(D) = Probability that it is labelled D

P(E) = Probability that it is labelled E

P(J) =

P(A) =

P(D) =

P(E) =

∴ P(N) = + + +

∴ P(N) =

(c) the probability P(S), that it is labelled with a letter that has at least one line of symmetry is;

P(S) = P(A) + P(B) + P(C) + P(D) + P(E) + P(H)

Where;

P(A) = Probability that it is labelled A

P(B) = Probability that it is labelled B

P(C) = Probability that it is labelled C

P(D) = Probability that it is labelled D

P(E) = Probability that it is labelled E

P(H) = Probability that it is labelled H

Cards with letters A, B, C, D, E and H are selected because these letters have at least one line of symmetry. A line of symmetry is a line that cuts an object into two identical halves. Letters A, B, C, D, E and H can each be cut into two identical halves.

P(A) =

P(B) =

P(C) =

P(D) =

P(E) =

P(H) =

∴ P(S) = + + + +

∴ P(S) =

congruent segments

step-by-step explanation: