I'm struggling on how can I solve the following problem. I tried with a lot of calculus without success, but i think that the way to solve is:

$$P(\text{at least one failure})=1−P(\text{no failures in both selections})$$

Here is the problem:

A manufacturer produces three types of lamps: A, B, and C. The probabilities of each type of lamp failing during testing are as follows:

$P(\text{A fails}) = 0.1$

$P(\text{B fails}) = 0.2$

$P(\text{C fails}) = 0.05$

A package contains 2 type A lamps, 3 type B lamps and 4 type C lamps. If two lamps are selected at random, what is the probability that at least one of them will fail, considering that the events are independents?

Anyone could give me a hint on how can I calculate this?

  • $\begingroup$ You probably need to add some independence assumption. $\endgroup$
    – Zhanxiong
    Apr 22 at 23:52
  • $\begingroup$ Ok @Zhanxiong, I edited. $\endgroup$
    – Ga13
    Apr 22 at 23:56
  • 2
    $\begingroup$ A hint that is always handy in these questions is whenever you see "at least" then it's code for "work out probability of 0, then 1 minus that." Usually. $\endgroup$
    – Huy Pham
    Apr 23 at 1:47

1 Answer 1


I believe the calculation is quite verbose, but the hint is applying the law of total probability, i.e., by decomposing the experiment into two steps. To begin with, you pick up two lamps. Next, you perform tests on these two selected lamps -- since the types of lamps picked in the previous step are available to us now, it is straightforward to apply the given condition in this step.

To convert this idea into mathematical formulas, let's introduce some notations. Let $T_i$ denote the type of lamp in the $i$-th pick, and $R_i$ be the test result for the $i$-th pick, $i = 1, 2$. The probability of your interest is $P(R_1 = S, R_2 = S)$ (where I use "$S$" to denote "success" and "$F$" to denote "failure"), i.e., the event of "no failures in both selections". By the law of total probability: \begin{align*} & P(R_1 = S, R_2 = S) \\ =& \sum_{i \in \{A, B, C\}}\sum_{j \in \{A, B, C\}} P(R_1 = S, R_2 = S|T_1 = i, T_2 = j)P(T_1 = i, T_2 = j). \end{align*} Clearly, there are nine summands in the right hand side of the above expression, and I will not spell out the details but just illustrate the evaluation of one of them as follows: \begin{align*} & P(R_1 = S, R_2 = S | T_1 = A, T_2 = A) = (1 - 0.1) \times (1 - 0.1) = 0.81, \\ & P(T_1 = A, T_2 = A) = \frac{\binom{2}{2}}{\binom{9}{2}} = \frac{1}{36}. \end{align*}


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