157. The battery fired six shots at the object. The probability of hitting the object with one shot is 0.3. Find: a) the most probable number of hits; b) the probability of the most probable number of hits; c) the probability that the object will be destroyed, if for this at least two hits are required.
Solution. According to the condition, $n=6 ; p=0.3 ; q=0.7$. a) Let's find the most probable number of hits using the formula

$$
n p-q<k_{0}<n p+p
$$

Substituting the data from the problem, we get

$$
6 \cdot 0.3 - 0.7 < k_{0} < 6 \cdot 0.3 + 0.3 \text { or } 1.1 < k_{0} < 2.1 \text { , }
$$

From this, $k_{0}=2$.

6) Let's find the probability of the most probable number of hits using the Bernoulli formula

$$
P_{6}(2)=C_{6}^{2} p^{2} q^{4}=\frac{6 \cdot 5}{1 \cdot 2} 0.3^{2} \cdot 0.7^{4}=0.324
$$

c) Let's find the probability that the object will be destroyed. According to the condition, for this, it is sufficient to have 2, 3, 4, 5, or 6 hits. These events are mutually exclusive, so the probability of the object being destroyed is equal to the sum of the probabilities of these events:

$$
P=P_{6}(2)+P_{6}(3)+P_{6}(4)+P_{6}(5)+P_{6}(6) .
$$

However, it is easier to first find the probability $Q$ of the opposite event (no hits or one hit):

$$
Q=P_{6}(0)+P_{6}(1)=q^{6}+C_{6}^{1} p q^{5}=0.7^{6}+6 \cdot 0.3 \cdot 0.7^{5}=0.42
$$

The desired probability that the object will be destroyed,

$$
P=1-Q=1-0.42=0.58
$$