Suppose you just received a shipment of eleven televisions. Two of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work?

Respuesta :

Answer: [tex]P(\text{ both the television are working })= 0.65454545 [/tex]

[tex]P(\text{ at least one the two televisions does not work }) =0.34545454545.[/tex]

Explanation:

Since the total number of television= 11

Number of defective television = 2

Remaining  television that are working = 9

 [tex]P(\text{ both the television are working }) = \frac{ ^9C_2}{^{11}C_2}\\P(\text{ both the television are working })=\frac{9\times 8}{11\times10}\\P(\text{ both the television are working })= \frac{72}{110}\\P(\text{both the television are working })= 0.65454545[/tex]

Now,

[tex]P(\text{ at least one the two televisions does not work }) = 1- \frac{72}{110}\\ P(\text{ at least one the two televisions does not work }) =\frac{110-72}{110}\\P(\text{ at least one the two televisions does not work }) = \frac{38}{110}\\P(\text{ at least one the two televisions does not work }) =0.34545454545.[/tex]

∴ [tex]P(\text{ both the television are working })= \frac{72}{110}[/tex]

[tex]P(\text{ both the television are working })= 0.65454545[/tex]

[tex]P(\text{ at least one the two televisions does not work }) =0.34545454545.[/tex]

The probability at least one of the two televisions does not​ work is 0.345

Given the following parameters:

total  number of television shipped = 11

Defective television = 2

Non defective = 11 - 2 = 9

  • Probability is the likelihood or chance that an event will occur.
  • Probability = Expected outcome/Total outcome

Pr(two television does not work) = [tex]\frac{9C2}{11C2}[/tex]

[tex]9C2 =\frac{9!}{7!2!}\\9C2 =\frac{9\times 8 \times 7!}{7!2!}\\9C2=\frac{72}{2} = 36[/tex]

Similarly;

[tex]11C2 =\frac{11!}{9!2!}\\11C2 =\frac{11\times 10 \times 9!}{9!2!}\\11C2=\frac{110}{2} = 55[/tex]

  • Pr(two television does not work) = [tex]\frac{36}{55}[/tex]

The probability at least one of the two televisions does not​ work = [tex]1-\frac{36}{55}[/tex]

The probability at least one of the two televisions does not​ work = [tex]\frac{19}{55} = 0.345[/tex]

Hence the probability at least one of the two televisions does not​ work is 0.345

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