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34kurt
3 years ago
11

One technique of killing cancer cells involves inserting microscopic synthetic rods, called carbon nanotubules, into the cell. W

hen the rods are exposed to near-infrared light from a laser, they heat up, killing the cell, while cells without rods are left unscathed (Wong et al., 2005). Suppose that five nanotubules are inserted in a single cancer cell. Independently of each other, they become exposed to near-infrared light with probabilities 0.2, 0.4, 0.3, 0.6, and 0.5. What is the probability that the cell will be killed
Mathematics
1 answer:
Vlad [161]3 years ago
4 0

Answer:

The probability that the cell will be killed is 0.9328.

Step-by-step explanation:

We are given that five nanotubules are inserted in a single cancer cell. Independently of each other, they become exposed to near-infrared light with probabilities 0.2, 0.4, 0.3, 0.6, and 0.5.

Let the event that a cell is killed be 'A' and the event where the ith nanotubule kill the cell be '\text{B}_i'.

This means that the cell will get killed if \text{B}_1 \bigcup \text{B}_2 \bigcup \text{B}_3 \bigcup \text{B}_4 \bigcup \text{B}_5 happens. This represents that the cell is killed if nanotubule 1 kills the cell, or nanotubule 2 kills the cell, and so on.

Here, P(\text{B}_1) = 0.2, P(\text{B}_2) = 0.4, P(\text{B}_3) = 0.3, P(\text{B}_4) = 0.6, P(\text{B}_5) = 0.5.

So, the probability that the cell will be killed is given by;

P(A)= 1 - [(1 - P(\text{B}_1)) \times (1 - P(\text{B}_2)) \times (1 - P(\text{B}_3)) \times (1 - P(\text{B}_4)) \times (1 - P(\text{B}_5))]

P(A) = 1 - [(1 - 0.2) \times (1 - 0.4) \times (1 - 0.3) \times (1 - 0.6) \times (1 - 0.5)]

P(A) = 1 - (0.8 \times 0.6 \times 0.7 \times 0.4 \times 0.5)

P(A) = 1 - 0.0672 = 0.9328

Hence, the probability that the cell will be killed is 0.9328.

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