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Feliz [49]
3 years ago
15

Maria and Farida had 250 beads altogether. After Maria used 18 beads to make a bracelet and Farida gave away 2/5 of her beads, t

hey have the same number of beads left. How many beads did Maria have at first?
Mathematics
1 answer:
sashaice [31]3 years ago
7 0
M + f = 250 ;
m - 18 = ( 3 / 5)f ;

5m / 5 - 90 / 5 = 3f;
5m - 90 = 3f ;
5m - 3f = 90;
But, f = 250 - m;
5m - 3( 250 - m ) = 90 ;
5m - 750 + 3m = 90 ;
8m = 840;
m = 840 ÷ 8 ;
m = 105( beads for Maria );
f = 250 - 105 ;
f = 145 ( beads for Farida ) ;


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NikAS [45]

Answer:

a) The expected number of goals team A will score is 5.1

b) The probability that team A will score a total of 5 goals is 0.1147

Step-by-step explanation:

Let X be the amount of goals scored by team A in both matches. Let X1 and X2 be the total amount of goals team A scores in match 1 and 2 respectively, then X = X1+X2, and also

a)

E(X) = E(X1+X2) = E(X1)+E(X2) = 0.6*2+0.4*3 + 0.3*2+0.7*3 = 5.1

b) In order for X to be equal to 5 we have 5 possibilities

- X1 is 0 and X2 is 5

- X1 is 1 and X2 is 4

- X1 is 2 and X2 is 3

- X1 is 3 and X2 is 2

- X1 is 4 and X2 is 1

- X1 is 5 and X2 is 0

Let T1 be a poisson distribution with mean λ = 2, then

P(T1=0) = e^{-2}

P(T1=1) = 2 * e^{-2}

P(T1=2) = 2 * e^{-2}

P(T1=3) = \frac{4}{3} \, e^{-2}

P(T1=4) = \frac{2}{3}\, e^{-2}

P(T1=5) = \frac{4}{15}\,e^{-2}

Lets do the same with a Poisson distribution T2 with mean λ = 3

P(T2=0) = e^{-3}

P(T2=1) = 3 \, e^{-3}\\P(T2=2) = \frac{9}{2} \, e^{-3}\\P(T2=3) = \frac{9}{2} \, e^{-3}\\P(T2=4) = \frac{27}{8} \, e^{-3}\\P(T2=5) = \frac{81}{40} \, e^{-3}

Now, we are ready to compute the probability that X is equal to 5.

P(X1 = 0, X2 = 5) = (0.6* e^{-2} + 0.4*e^{-3}) * (0.3*\frac{4}{15}e^{-2}  + 0.7*\frac{81}{40} e^{-3}) = 0.00823\\P(X1 = 1, X2 = 4) = (0.6* 2e^{-2} + 0.4*3 e^{-3}) * (0.3*\frac{2}{3}e^{-2}  + 0.7*\frac{27}{8} e^{-3}) = 0.03214\\P(X1 = 2, X2 = 3) = (0.6* 2e^{-2} + 0.4*\frac{9}{2}e^{-3}) * (0.3*\frac{4}{3}e^{-2}  + 0.7*\frac{9}{2} e^{-3}) = 0.05317\\P(X1 = 3, X2 = 2) = (0.6* \frac{4}{3}e^{-2} + 0.4*\frac{9}{2}*e^{-3}) * (0.3*2e^{-2}  + 0.7*\frac{9}{2} e^{-3}) = 0.0471

P(X1 = 4, X2 = 1) = (0.6* \frac{2}{3}e^{-2} + 0.4*\frac{27}{8}e^{-3}) * (0.3*2e^{-2}  + 0.7*3e^{-3}) = 0.0225\\P(X1 = 5, X2 = 0) = (0.6* \frac{4}{15}e^{-2} + 0.4*\frac{81}{40}e^{-3}) * (0.3*e^{-2}  + 0.7*e^{-3}) = 0.0047

We can conclude that

P(X = 5) = 0.00823+0.03214+0.05317+0.0471+0.0225+0.0047 = 0.1147

The probability that team A will score a total of 5 goals is 0.1147

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