Chris has a bag of sweets. there are more than 20 sweets in a bag.
he shares his sweets equally between six people.
chris says "the number of sweets in the packet was prime "
is he correct? cirle: yes or no.
explain your answer below

The answer is no because if it is over 20 but he shared is equally with 6 people then it stands to reason that the amount was exactly 24 which isnt a prime number

241,2.1Pg51) Sample Spacedenoted by , is the set of all possible outcomes of that experiment.(241,2.1Pg52i) Eventany collection (subset) of outcomes contained in the sample space . An -Term- is said to be simple if it consists of exactly one outcome and compound if it consists of more than one outcome.(241,2.1Pg52ii) Simple Eventan event that consists of exactly one outcome.(241,2.1Pg52iii) Compound Eventan event that consists of more than one outcome.(241,2.1Pg53) UnionThe -Term- of two events A and B, denoted A ∪ B and read " A or B," is the event consisting of all outcomes that are either in A or in B or in both events, that is, all outcomes in at least one of the events.(241,2.1Pg53) IntersectionThe -Term- of two events A and B, denoted by A ∩ B and read "A and B," is the event consisting of all outcomes that are in both A and B.(241,2.1Pg53) ComplementThe -Term- of an event A, denoted by A', is the set of all outcomes in that are not contained in A.(241,2.1Pg53) Mutually ExclusiveWhen A and B have no outcomes in common, they are said to be -Blank- or -Term- events. Mathematicians write this compactly as A ∩ B = ∅ . (241,2.1Pg53) Disjoint; When A and B have no outcomes in common, they are said to be -Term- or -Black- events. Mathematicians write this compactly as A ∩ B = ∅(241,2.1Pg53) ∅denotes the event consisting of no outcomes whatsoever.(null or empty event)( 241,2.2Pg57) Axioms of probabilityAxiom 1: For any event A, P(A) >= 0. Axiom 2: P() = 1. Axiom 3: If A1, A2, A3, ... is an infinite collection of disjoint events, then P(A1 ∪ A2 ∪ A3 ...) = ∑^∞(i=1) P(Ai)(241,2.2Pg57) P(∅)0(241,2.2Pg60) P(A')1 - P(A)(241,2.2Pg60) P(A)'s maximum valueP(A) <= 1(241,2.2Pg60) P(A ∪ B)P(A) + P(B) - P(A ∩ B)(241,2.2Pg62) P( A ∪ B ∪ C)P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)(241,2.3Pg66) P(A) for equally likely eventsP(A) = (N(A))/N (N = # of outcomes in , N(A) = # of outcomes in A)(241,2.3Pg69) PermutationAny ordered sequence of k objects taken from a set of n distinct objects is called a -Term- of size k of the objects. The number of -Terms- of size k that can be constructed from n objects is denoted by nPk(241,2.3Pg69) FactorialFor any positive integer m, m! is read " m -Term-" and is defined by m! = m(m-1)(m-2)...(2)(1).(241,2.3Pg69) 0!1(241,2.3Pg69) nPk = n!/((n-k)!)...(241,2.3Pg70) CombinationGiven a set of n disjoint objects, any unordered subset of size k of the objects is called a -Term-. The number of -Terms- of size k that can be formed from n distinct objects will be donated by nCk.(241,2.3Pg70) nCk(nPk)/k! = n!/(k!(n - k)!)(241,2.4Pg75) Conditional probability of A given that B has occurredFor any two events A and B with P(B) > 0, the -Term- is defined by P(A | B) =P(A ∩ B)/P(B)(241,2.4Pg75) P(A | B)P(A ∩ B)/P(B)(241,2.4Pg77) The Multiplication Rule P(A ∩ B)P(A | B) x P(B)(241,2.4Pg79) The Law of Total ProbabilityLet A1, ... , Ak be mutually exclusive and exhaustive events. Then for any other event B. P(B) = P(B | A1) x P(A1) + ... + P(B | Ak) x P(Ak) = ∑^k(i=1) P(B | Ai)P(Ai)(241,2.4Pg80) Bayes' TheoremLet A1, ... , Ak be a collection of mutually exclusive and exhaustive events with P(Ai) > 0 for i=1, ... , k. Then for any other event B, for which P(B)>0: P(Aj | B) = (P(Aj ∩ B))/P(B) = [P(B | Aj)P(Aj)]/[∑^k(i=1) P(B | Ai)P(Ai)] for j = 1, ... , k(241,2.5Pg85i) IndependentTwo events A and B are -Term- if P(A | B) = P(A) and are dependent otherwise.(241,2.5Pg85ii) DependentTwo events A and B are -Term- if P(A | B) =/= P(A) and are independent otherwise.(241,2.5Pg86) A and B are independent iffP(A ∩ B) = P(A) x P(B)(241,2.5Pg87) Mutually IndependentEvents A1, ..., An are mutually independent if for every k(k = 2, 3, ..., n) and every subset of indices i1, i2, ..., ik, P(Ai1 ∩ Ai2 ∩ ... ∩ Aik) = P(Ai1)xP(Ai2)x ... P(Aik)