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yanalaym [24]
3 years ago
14

Let A={1'3'5'6} B={0'7} check whether A and B are disjoint sets

Mathematics
1 answer:
Setler79 [48]3 years ago
4 0

Given:

The sets are A=\{1,3,5,6\} and B=\{0,7\}.

To find:

Whether A and B are disjoint sets or not.

Solution:

Two sets are called disjoint sets if there are no common elements between them.

We have,

A=\{1,3,5,6\}

B=\{0,7\}

Clearly, there is not common elements between the set A and set B.

Therefore, the sets A and B are disjoint sets.

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Laura and Philip each fire one shot at a target. Laura has probability 0.4 of hitting the target, and Philip has probability 0.1
Pepsi [2]

Answer:

(i) 0.46, (ii)0.42, and (iii)0.143

Step-by-step explanation:

Let p be the probability of hitting the target and q be the probability of missing the target by Laura.

Given that p=0.4\;\cdots (1)

As Laura either hit or miss the target, so p+q=1.

\Rightarrow q=1-p=0.6 \; \cdots (2)

Again, let r be the probability of hitting the target and s be the probability of missing the target by Philip.

Here, r=0.1\;\cdots (3)

Similarly, \Rightarrow s=1-r=0.9\;\cdots (4)

(i)The probability that the target is hit means that the target is not missed by both, either of one or both hit the target.

=1-(probability of missing the target by both)

=1-qr [from equation (2) and (4) ]

=1-0.6\times 0.9

=1-0.54

=0.46

(ii) The probability that the target is hit by exactly one shot means either of one hit the target.

=Hit by Laura and missed by Philip or hit by Philip and missed by Laura

=0.4\times 0.9+0.1\times 0.6 [from equations (1),(4) and (3),(2)]

=0.36+0.06

=0.42

(iii)Given that the target was hit by exactly one shot, so, the given probability is 0.42 [from (ii) part]

No, the probability that the target was hit by Philip = probability of hitting the target by Philip and missing the target by Laura

=0.1\times 0.6 [from equations (3) and (2)]

=0.06

So, the probability of hitting the target by Philip

=\frac {0.06}{0.42}

=\frac {1}{7}

=0.143 (approx)

6 0
3 years ago
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