Question

An elementary school is offering 2 language classes: one in Spanish (S) and one in French (F). Given that P(S) = 50%, P(F) = 40%, P(S ∪ F) = 70%, find the probability that a randomly selected student (a) is taking Spanish given that he or she is taking French; (b) is not taking French given that he or she is not taking Spanish. 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears.

Answer 1

Answer:

(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French = 0.5 .

(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish = 0.6 .

Step-by-step explanation:

We are given that an elementary school is offering 2 language classes ;

 Spanish Language is denoted by S and French language is denoted by F.

Also we are given, P(S) = 0.5 {Probability of students taking Spanish language}

P(F) = 0.4 {Probability of students taking French language}

$P(S\bigcup F)$ = 0.7 {Probability of students taking Spanish or French Language}

We know that,  $P(A\bigcup B)$  = $P(A) + P(B) -$ $P(A\bigcap B)$

So, $P(S\bigcap F)$ = $P(S) + P(F) - P(S\bigcup F)$ = 0.5 + 0.4 - 0.7 = 0.2

$P(S\bigcap F)$ means Probability of students taking  both Spanish and French Language.

Also, P(S)' = 1 - P(S) = 1 - 0.5 = 0.5

         P(F)' = 1 - P(F) = 1 - 0.4 = 0.6

        $P(S'\bigcap F')$ = 1 -  $P(S\bigcup F)$ = 1 - 0.7 = 0.3

(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French is given by P(S/F);

  P(S/F) = $\frac{P(S\bigcap F)}{P(F)}$ = $\frac{0.2}{0.4}$ = 0.5

(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish is given by P(F'/S');

   P(F'/S') = $\frac{P(S'\bigcap F')}{P(S')}$ = $\frac{1- P(S\bigcup F)}{1-P(S)}$ = $\frac{0.3}{0.5}$ = 0.6 .

Note: 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears  ; This question is incomplete please provide with complete detail.